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Sec on 4.4
Curve Sketching
  V63.0121.001: Calculus I
Professor Ma hew Leingang
       New York University


      April 13, 2011
                             .
Announcements
   Quiz 4 on Sec ons 3.3,
   3.4, 3.5, and 3.7 this
   week (April 14/15)
   Quiz 5 on Sec ons
   4.1–4.4 April 28/29
   Final Exam Thursday May
   12, 2:00–3:50pm
   I am teaching Calc II MW
   2:00pm and Calc III TR
   2:00pm both Fall ’11 and
   Spring ’12
Objectives

   given a func on, graph it
   completely, indica ng
       zeroes (if easy)
       asymptotes if applicable
       cri cal points
       local/global max/min
       inflec on points
Why?

Graphing func ons is like
dissec on
Why?

Graphing func ons is like
dissec on … or diagramming
sentences
Why?

Graphing func ons is like
dissec on … or diagramming
sentences
You can really know a lot
about a func on when you
know all of its anatomy.
The Increasing/Decreasing Test
 Theorem (The Increasing/Decreasing Test)
 If f′ > 0 on (a, b), then f is increasing on (a, b). If f′ < 0 on (a, b),
 then f is decreasing on (a, b).

 Example
                                                                f(x)
  f(x) = x3 + x2
                                                    .
The Increasing/Decreasing Test
 Theorem (The Increasing/Decreasing Test)
 If f′ > 0 on (a, b), then f is increasing on (a, b). If f′ < 0 on (a, b),
 then f is decreasing on (a, b).

 Example
                                                                   f(x)
                                                          f′ (x)
   f(x) = x3 + x2
  f′ (x) = 3x2 + 2x                                 .
Testing for Concavity
 Theorem (Concavity Test)
 If f′′ (x) > 0 for all x in (a, b), then the graph of f is concave upward
 on (a, b) If f′′ (x) < 0 for all x in (a, b), then the graph of f is concave
 downward on (a, b).

 Example
                                                             f(x)

   f(x) = x3 + x2
                                                    .
Testing for Concavity
 Theorem (Concavity Test)
 If f′′ (x) > 0 for all x in (a, b), then the graph of f is concave upward
 on (a, b) If f′′ (x) < 0 for all x in (a, b), then the graph of f is concave
 downward on (a, b).

 Example
                                                        f′ (x) f(x)

   f(x) = x3 + x2
  f′ (x) = 3x2 + 2x                                 .
Testing for Concavity
 Theorem (Concavity Test)
 If f′′ (x) > 0 for all x in (a, b), then the graph of f is concave upward
 on (a, b) If f′′ (x) < 0 for all x in (a, b), then the graph of f is concave
 downward on (a, b).

 Example
                                                 f′′ (x) f′ (x) f(x)

    f(x) = x3 + x2
   f′ (x) = 3x2 + 2x                                .
  f′′ (x) = 6x + 2
Graphing Checklist
To graph a func on f, follow this plan:
 0. Find when f is posi ve, nega ve, zero,
    not defined.
Graphing Checklist
To graph a func on f, follow this plan:
 0. Find when f is posi ve, nega ve, zero,
    not defined.
 1. Find f′ and form its sign chart.
    Conclude informa on about
    increasing/decreasing and local
    max/min.
Graphing Checklist
To graph a func on f, follow this plan:
 0. Find when f is posi ve, nega ve, zero,
    not defined.
 1. Find f′ and form its sign chart.
    Conclude informa on about
    increasing/decreasing and local
    max/min.
 2. Find f′′ and form its sign chart.
    Conclude concave up/concave down
    and inflec on.
Graphing Checklist
To graph a func on f, follow this plan:
 3. Put together a big chart to assemble
    monotonicity and concavity data
Graphing Checklist
To graph a func on f, follow this plan:
 3. Put together a big chart to assemble
    monotonicity and concavity data
 4. Graph!
Outline
 Simple examples
    A cubic func on
    A quar c func on

 More Examples
   Points of nondifferen ability
   Horizontal asymptotes
   Ver cal asymptotes
   Trigonometric and polynomial together
   Logarithmic
Graphing a cubic
 Example
 Graph f(x) = 2x3 − 3x2 − 12x.
Graphing a cubic
 Example
 Graph f(x) = 2x3 − 3x2 − 12x.
 (Step 0) First, let’s find the zeros. We can at least factor out one
 power of x:
                          f(x) = x(2x2 − 3x − 12)
 so f(0) = 0. The other factor is a quadra c, so we the other two
 roots are
                       √                          √
                   3 ± 32 − 4(2)(−12) 3 ± 105
               x=                          =
                              4                   4
 It’s OK to skip this step for now since the roots are so complicated.
Step 1: Monotonicity
                f(x) = 2x3 − 3x2 − 12x
            =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2)
 We can form a sign chart from this:

                           .
Step 1: Monotonicity
                f(x) = 2x3 − 3x2 − 12x
            =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2)
 We can form a sign chart from this:

                           .                x−2
                                       2
                                            x+1
                    −1                      f′ (x)
                    −1                 2    f(x)
Step 1: Monotonicity
                f(x) = 2x3 − 3x2 − 12x
            =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2)
 We can form a sign chart from this:
                  −        .−              +
                                               x−2
                                       2
                                               x+1
                    −1                         f′ (x)
                    −1                 2       f(x)
Step 1: Monotonicity
                f(x) = 2x3 − 3x2 − 12x
            =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2)
 We can form a sign chart from this:
                  −        .−              +
                                               x−2
                                       2
                  −         +              +
                                               x+1
                   −1                          f′ (x)
                    −1                 2       f(x)
Step 1: Monotonicity
                f(x) = 2x3 − 3x2 − 12x
            =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2)
 We can form a sign chart from this:
                  −        .−              +
                                               x−2
                                       2
                  −         +              +
                                               x+1
                   −1                          f′ (x)
                  +
                   −1                  2       f(x)
Step 1: Monotonicity
                f(x) = 2x3 − 3x2 − 12x
            =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2)
 We can form a sign chart from this:
                  −        .−              +
                                               x−2
                                       2
                  −         +              +
                                               x+1
                   −1                          f′ (x)
                  +         −
                   −1                  2       f(x)
Step 1: Monotonicity
                f(x) = 2x3 − 3x2 − 12x
            =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2)
 We can form a sign chart from this:
                  −        .−              +
                                               x−2
                                       2
                  −         +              +
                                               x+1
                   −1                          f′ (x)
                  +         −              +
                   −1                  2       f(x)
Step 1: Monotonicity
                f(x) = 2x3 − 3x2 − 12x
            =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2)
 We can form a sign chart from this:
                  −        .−              +
                                               x−2
                                       2
                  −         +              +
                                               x+1
                   −1                          f′ (x)
                  +         −              +
                  ↗−1                  2       f(x)
Step 1: Monotonicity
                f(x) = 2x3 − 3x2 − 12x
            =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2)
 We can form a sign chart from this:
                  −        .−              +
                                               x−2
                                       2
                  −         +              +
                                               x+1
                   −1                          f′ (x)
                  +         −              +
                  ↗−1       ↘          2       f(x)
Step 1: Monotonicity
                f(x) = 2x3 − 3x2 − 12x
            =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2)
 We can form a sign chart from this:
                  −        .−              +
                                               x−2
                                       2
                  −         +              +
                                               x+1
                   −1                          f′ (x)
                  +         −              +
                  ↗−1       ↘          2   ↗   f(x)
Step 1: Monotonicity
                f(x) = 2x3 − 3x2 − 12x
            =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2)
 We can form a sign chart from this:
                  −        .−              +
                                               x−2
                                       2
                  −         +              +
                                               x+1
                   −1                          f′ (x)
                  +         −              +
                  ↗−1       ↘          2   ↗   f(x)
                   max
Step 1: Monotonicity
                f(x) = 2x3 − 3x2 − 12x
            =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2)
 We can form a sign chart from this:
                  −        .−              +
                                               x−2
                                       2
                  −         +              +
                                               x+1
                   −1                          f′ (x)
                  +         −              +
                  ↗−1       ↘      2       ↗   f(x)
                   max            min
Step 2: Concavity

                      f′ (x) = 6x2 − 6x − 12
                  =⇒ f′′ (x) = 12x − 6 = 6(2x − 1)

 Another sign chart:
                          .
Step 2: Concavity

                      f′ (x) = 6x2 − 6x − 12
                  =⇒ f′′ (x) = 12x − 6 = 6(2x − 1)

 Another sign chart:
                          .
                                             f′′ (x)
                              1/2            f(x)
Step 2: Concavity

                      f′ (x) = 6x2 − 6x − 12
                  =⇒ f′′ (x) = 12x − 6 = 6(2x − 1)

 Another sign chart:
                            .
                       −−                    f′′ (x)
                                1/2          f(x)
Step 2: Concavity

                      f′ (x) = 6x2 − 6x − 12
                  =⇒ f′′ (x) = 12x − 6 = 6(2x − 1)

 Another sign chart:
                            .
                       −−             ++     f′′ (x)
                                1/2          f(x)
Step 2: Concavity

                      f′ (x) = 6x2 − 6x − 12
                  =⇒ f′′ (x) = 12x − 6 = 6(2x − 1)

 Another sign chart:
                            .
                       −−             ++     f′′ (x)
                       ⌢        1/2          f(x)
Step 2: Concavity

                      f′ (x) = 6x2 − 6x − 12
                  =⇒ f′′ (x) = 12x − 6 = 6(2x − 1)

 Another sign chart:
                            .
                       −−           ++       f′′ (x)
                       ⌢        1/2 ⌣        f(x)
Step 2: Concavity

                      f′ (x) = 6x2 − 6x − 12
                  =⇒ f′′ (x) = 12x − 6 = 6(2x − 1)

 Another sign chart:
                            .
                       −−           ++       f′′ (x)
                       ⌢        1/2 ⌣        f(x)
                                 IP
Step 3: One sign chart to rule them all
 Remember, f(x) = 2x3 − 3x2 − 12x.

                     .
Step 3: One sign chart to rule them all
 Remember, f(x) = 2x3 − 3x2 − 12x.

             +   −.      −       +   f′ (x)
             ↗−1 ↘       ↘ 2     ↗   monotonicity
Step 3: One sign chart to rule them all
 Remember, f(x) = 2x3 − 3x2 − 12x.

             +  −.  −   +            f′ (x)
            ↗−1 ↘   ↘ 2 ↗            monotonicity
            −− −− ++    ++           f′′ (x)
            ⌢ ⌢ 1/2 ⌣   ⌣            concavity
Step 3: One sign chart to rule them all
 Remember, f(x) = 2x3 − 3x2 − 12x.

             +  −.  −   +            f′ (x)
            ↗−1 ↘   ↘ 2 ↗            monotonicity
            −− −− ++    ++           f′′ (x)
            ⌢ ⌢ 1/2 ⌣   ⌣            concavity
               7  −61/2 −20          f(x)
              −1   1/2   2           shape of f
              max   IP  min
monotonicity and concavity


             II        I
                   .

             III       IV
monotonicity and concavity
                            decreasing,
                            concave
                            down
             II        I
                   .

             III       IV
monotonicity and concavity
         increasing,              decreasing,
         concave                  concave
         down                     down
                   II        I
                         .

                   III       IV
monotonicity and concavity
         increasing,             decreasing,
         concave                 concave
         down                    down
                   II       I
                        .

                  III       IV
         decreasing,
         concave
         up
monotonicity and concavity
         increasing,             decreasing,
         concave                 concave
         down                    down
                   II       I
                        .

                  III       IV
         decreasing,             increasing,
         concave                 concave
         up                      up
Step 3: One sign chart to rule them all
 Remember, f(x) = 2x3 − 3x2 − 12x.

             +  −.  −   +            f′ (x)
            ↗−1 ↘   ↘ 2 ↗            monotonicity
            −− −− ++    ++           f′′ (x)
            ⌢ ⌢ 1/2 ⌣   ⌣            concavity
               7  −61/2 −20          f(x)
              −1   1/2   2           shape of f
              max   IP  min
Step 3: One sign chart to rule them all
 Remember, f(x) = 2x3 − 3x2 − 12x.

             +  −.  −   +            f′ (x)
            ↗−1 ↘   ↘ 2 ↗            monotonicity
            −− −− ++    ++           f′′ (x)
            ⌢ ⌢ 1/2 ⌣   ⌣            concavity
               7  −61/2 −20          f(x)
              −1   1/2   2           shape of f
              max   IP  min
Step 3: One sign chart to rule them all
 Remember, f(x) = 2x3 − 3x2 − 12x.

             +  −.  −   +            f′ (x)
            ↗−1 ↘   ↘ 2 ↗            monotonicity
            −− −− ++    ++           f′′ (x)
            ⌢ ⌢ 1/2 ⌣   ⌣            concavity
               7  −61/2 −20          f(x)
              −1   1/2   2           shape of f
              max   IP  min
Step 3: One sign chart to rule them all
 Remember, f(x) = 2x3 − 3x2 − 12x.

             +  −.  −   +            f′ (x)
            ↗−1 ↘   ↘ 2 ↗            monotonicity
            −− −− ++    ++           f′′ (x)
            ⌢ ⌢ 1/2 ⌣   ⌣            concavity
               7  −61/2 −20          f(x)
              −1   1/2   2           shape of f
              max   IP  min
f(x)
Step 4: Graph
       f(x) = 2x3 − 3x2 − 12x
       ( √       ) (−1, 7)
        3− 105
           4   ,0            (0, 0)
                           .                  ( x√         )
                               (1/2, −61/2)
                                               3+ 105
                                                 4    ,0
                                 (2, −20)

                      7  −61/2 −20              f(x)
                     −1   1/2   2               shape of f
                     max   IP  min
f(x)
Step 4: Graph
       f(x) = 2x3 − 3x2 − 12x
       ( √       ) (−1, 7)
        3− 105
           4   ,0            (0, 0)
                           .                  ( x√         )
                               (1/2, −61/2)
                                               3+ 105
                                                 4    ,0
                                 (2, −20)

                      7  −61/2 −20              f(x)
                     −1   1/2   2               shape of f
                     max   IP  min
f(x)
Step 4: Graph
       f(x) = 2x3 − 3x2 − 12x
       ( √       ) (−1, 7)
        3− 105
           4   ,0            (0, 0)
                           .                  ( x√         )
                               (1/2, −61/2)
                                               3+ 105
                                                 4    ,0
                                 (2, −20)

                      7  −61/2 −20              f(x)
                     −1   1/2   2               shape of f
                     max   IP  min
f(x)
Step 4: Graph
       f(x) = 2x3 − 3x2 − 12x
       ( √       ) (−1, 7)
        3− 105
           4   ,0            (0, 0)
                           .                  ( x√         )
                               (1/2, −61/2)
                                               3+ 105
                                                 4    ,0
                                 (2, −20)

                      7  −61/2 −20              f(x)
                     −1   1/2   2               shape of f
                     max   IP  min
f(x)
Step 4: Graph
       f(x) = 2x3 − 3x2 − 12x
       ( √       ) (−1, 7)
        3− 105
           4   ,0            (0, 0)
                           .                  ( x√         )
                               (1/2, −61/2)
                                               3+ 105
                                                 4    ,0
                                 (2, −20)

                      7  −61/2 −20              f(x)
                     −1   1/2   2               shape of f
                     max   IP  min
Graphing a quartic

 Example
 Graph f(x) = x4 − 4x3 + 10
Graphing a quartic

 Example
 Graph f(x) = x4 − 4x3 + 10
 (Step 0) We know f(0) = 10 and lim f(x) = +∞. Not too many
                                 x→±∞
 other points on the graph are evident.
Step 1: Monotonicity
            f(x) = x4 − 4x3 + 10
        =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3)
Step 1: Monotonicity
                    f(x) = x4 − 4x3 + 10
                =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3)
 We make its sign chart.

                       .
Step 1: Monotonicity
                    f(x) = x4 − 4x3 + 10
                =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3)
 We make its sign chart.
                      0.                 4x2
                      0
Step 1: Monotonicity
                    f(x) = x4 − 4x3 + 10
                =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3)
 We make its sign chart.
                   +0.                   4x2
                    0
Step 1: Monotonicity
                    f(x) = x4 − 4x3 + 10
                =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3)
 We make its sign chart.
                   +0.      +
                                         4x2
                    0
Step 1: Monotonicity
                    f(x) = x4 − 4x3 + 10
                =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3)
 We make its sign chart.
                   +0.      +        +
                                         4x2
                    0
Step 1: Monotonicity
                    f(x) = x4 − 4x3 + 10
                =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3)
 We make its sign chart.
                   +0.      +          +
                                           4x2
                    0
                                   0
                                           (x − 3)
                                   3
Step 1: Monotonicity
                    f(x) = x4 − 4x3 + 10
                =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3)
 We make its sign chart.
                   +0 .     +          +
                                           4x2
                     0
                   −               0
                                           (x − 3)
                                   3
Step 1: Monotonicity
                    f(x) = x4 − 4x3 + 10
                =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3)
 We make its sign chart.
                   +0 .     +          +
                                           4x2
                     0
                   −        −      0
                                           (x − 3)
                                   3
Step 1: Monotonicity
                    f(x) = x4 − 4x3 + 10
                =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3)
 We make its sign chart.
                   +0 .     +        +
                                         4x2
                     0
                   −        −      0+
                                      (x − 3)
                                   3
Step 1: Monotonicity
                    f(x) = x4 − 4x3 + 10
                =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3)
 We make its sign chart.
                   +0 .     +        +
                                         4x2
                     0
                   −        −      0+
                                      (x − 3)
                                   3  f′ (x)
                      0            0
                      0            3  f(x)
Step 1: Monotonicity
                    f(x) = x4 − 4x3 + 10
                =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3)
 We make its sign chart.
                   +0 .     +        +
                                         4x2
                     0
                   −        −      0+
                                      (x − 3)
                                   3  f′ (x)
                   −0              0
                    0              3  f(x)
Step 1: Monotonicity
                    f(x) = x4 − 4x3 + 10
                =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3)
 We make its sign chart.
                   +0 .     +        +
                                         4x2
                     0
                   −        −      0+
                                      (x − 3)
                                   3  f′ (x)
                   −0       −      0
                    0              3  f(x)
Step 1: Monotonicity
                    f(x) = x4 − 4x3 + 10
                =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3)
 We make its sign chart.
                   +0 .     +        +
                                         4x2
                     0
                   −        −      0+
                                       (x − 3)
                                   3    ′
                   −0       −      0 + f (x)
                    0              3   f(x)
Step 1: Monotonicity
                    f(x) = x4 − 4x3 + 10
                =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3)
 We make its sign chart.
                   +0 .     +        +
                                         4x2
                     0
                   −        −      0+
                                       (x − 3)
                                   3    ′
                   −0       −      0 + f (x)
                   ↘0              3   f(x)
Step 1: Monotonicity
                    f(x) = x4 − 4x3 + 10
                =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3)
 We make its sign chart.
                   +0 .     +        +
                                         4x2
                     0
                   −        −      0+
                                       (x − 3)
                                   3    ′
                   −0       −      0 + f (x)
                   ↘0       ↘      3   f(x)
Step 1: Monotonicity
                    f(x) = x4 − 4x3 + 10
                =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3)
 We make its sign chart.
                   +0 .     +        +
                                         4x2
                     0
                   −        −      0+
                                       (x − 3)
                                   3    ′
                   −0       −      0 + f (x)
                   ↘0       ↘      3 ↗ f(x)
Step 1: Monotonicity
                    f(x) = x4 − 4x3 + 10
                =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3)
 We make its sign chart.
                   +0 .     +        +
                                         4x2
                     0
                   −        −      0+
                                       (x − 3)
                                   3    ′
                   −0       −      0 + f (x)
                   ↘0       ↘      3 ↗ f(x)
                                  min
Step 2: Concavity
            f′ (x) = 4x3 − 12x2
        =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)


               .
Step 2: Concavity
                     f′ (x) = 4x3 − 12x2
                 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)
 Here is its sign chart:

                           .
Step 2: Concavity
                     f′ (x) = 4x3 − 12x2
                 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)
 Here is its sign chart:
                           0
                           .               12x
                           0
Step 2: Concavity
                     f′ (x) = 4x3 − 12x2
                 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)
 Here is its sign chart:
                       −0
                        .                  12x
                        0
Step 2: Concavity
                     f′ (x) = 4x3 − 12x2
                 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)
 Here is its sign chart:
                       −0
                        .   +
                                           12x
                        0
Step 2: Concavity
                     f′ (x) = 4x3 − 12x2
                 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)
 Here is its sign chart:
                       −0
                        .   +        +
                                           12x
                        0
Step 2: Concavity
                     f′ (x) = 4x3 − 12x2
                 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)
 Here is its sign chart:
                       −0
                        .   +        +
                                           12x
                        0
                                 0
                                           x−2
                                 2
Step 2: Concavity
                     f′ (x) = 4x3 − 12x2
                 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)
 Here is its sign chart:
                       −0.   +       +
                                           12x
                         0
                       −         0
                                           x−2
                                 2
Step 2: Concavity
                     f′ (x) = 4x3 − 12x2
                 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)
 Here is its sign chart:
                       −0.   +       +
                                           12x
                         0
                       −     −   0
                                           x−2
                                 2
Step 2: Concavity
                     f′ (x) = 4x3 − 12x2
                 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)
 Here is its sign chart:
                       −0.   +       +
                                           12x
                         0
                       −     −   0   +
                                           x−2
                                 2
Step 2: Concavity
                     f′ (x) = 4x3 − 12x2
                 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)
 Here is its sign chart:
                       −0.     +       +
                                           12x
                         0
                       −       −   0   +
                                           x−2
                                   2       f′′ (x)
                           0       0
                           0       2       f(x)
Step 2: Concavity
                     f′ (x) = 4x3 − 12x2
                 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)
 Here is its sign chart:
                       −0.   +       +
                                           12x
                         0
                       −     −   0   +
                                           x−2
                                 2         f′′ (x)
                      ++0        0
                        0        2         f(x)
Step 2: Concavity
                     f′ (x) = 4x3 − 12x2
                 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)
 Here is its sign chart:
                       −0.   +       +
                                           12x
                         0
                       −     −
                             0       +
                                           x−2
                             2             f′′ (x)
                      ++0 −− 0
                        0    2             f(x)
Step 2: Concavity
                     f′ (x) = 4x3 − 12x2
                 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)
 Here is its sign chart:
                       −0.   +       +
                                           12x
                         0
                       −     −
                             0 +
                                           x−2
                             2             f′′ (x)
                      ++0 −− 0 ++
                        0    2             f(x)
Step 2: Concavity
                     f′ (x) = 4x3 − 12x2
                 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)
 Here is its sign chart:
                       −0.   +       +
                                           12x
                         0
                       −     −
                             0 +
                                           x−2
                             2             f′′ (x)
                      ++0 −− 0 ++
                      ⌣0     2             f(x)
Step 2: Concavity
                     f′ (x) = 4x3 − 12x2
                 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)
 Here is its sign chart:
                       −0.   +       +
                                           12x
                         0
                       −     −
                             0 +
                                           x−2
                             2             f′′ (x)
                      ++0 −− 0 ++
                      ⌣0 ⌢ 2               f(x)
Step 2: Concavity
                     f′ (x) = 4x3 − 12x2
                 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)
 Here is its sign chart:
                       −0.   +       +
                                           12x
                         0
                       −     −
                             0 +
                                           x−2
                             2             f′′ (x)
                      ++0 −− 0 ++
                      ⌣0 ⌢ 2 ⌣             f(x)
Step 2: Concavity
                     f′ (x) = 4x3 − 12x2
                 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)
 Here is its sign chart:
                       −0.   +       +
                                           12x
                         0
                       −     −
                             0 +
                                           x−2
                             2             f′′ (x)
                      ++0 −− 0 ++
                      ⌣0 ⌢ 2 ⌣             f(x)
                        IP
Step 2: Concavity
                     f′ (x) = 4x3 − 12x2
                 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)
 Here is its sign chart:
                       −0.   +       +
                                           12x
                         0
                       −     −0 +
                                           x−2
                              2            f′′ (x)
                      ++0 −− 0 ++
                      ⌣0 ⌢ 2 ⌣             f(x)
                        IP   IP
Step 3: Grand Unified Sign Chart
                    .
 Remember, f(x) = x − 4x3 + 10.
                   4



                −0 −     −0+      f′ (x)
                ↘0 ↘     ↘3↗      monotonicity
                                  f′′ (x)
                ++0 −− 0++ ++
                ⌣0 ⌢ 2⌣ ⌣         concavity
                  10   −6 −17     f(x)
                   0    2 3       shape
                  IP   IP min
Step 3: Grand Unified Sign Chart
                    .
 Remember, f(x) = x − 4x3 + 10.
                   4



                −0 −     −0+      f′ (x)
                ↘0 ↘     ↘3↗      monotonicity
                                  f′′ (x)
                ++0 −− 0++ ++
                ⌣0 ⌢ 2⌣ ⌣         concavity
                  10   −6 −17     f(x)
                   0    2 3       shape
                  IP   IP min
Step 3: Grand Unified Sign Chart
                    .
 Remember, f(x) = x − 4x3 + 10.
                   4



                −0 −     −0+      f′ (x)
                ↘0 ↘     ↘3↗      monotonicity
                                  f′′ (x)
                ++0 −− 0++ ++
                ⌣0 ⌢ 2⌣ ⌣         concavity
                  10   −6 −17     f(x)
                   0    2 3       shape
                  IP   IP min
Step 3: Grand Unified Sign Chart
                    .
 Remember, f(x) = x − 4x3 + 10.
                   4



                −0 −     −0+      f′ (x)
                ↘0 ↘     ↘3↗      monotonicity
                                  f′′ (x)
                ++0 −− 0++ ++
                ⌣0 ⌢ 2⌣ ⌣         concavity
                  10   −6 −17     f(x)
                   0    2 3       shape
                  IP   IP min
Step 3: Grand Unified Sign Chart
                    .
 Remember, f(x) = x − 4x3 + 10.
                   4



                −0 −     −0+      f′ (x)
                ↘0 ↘     ↘3↗      monotonicity
                                  f′′ (x)
                ++0 −− 0++ ++
                ⌣0 ⌢ 2⌣ ⌣         concavity
                  10   −6 −17     f(x)
                   0    2 3       shape
                  IP   IP min
y
Step 4: Graph

     f(x) = x4 − 4x3 + 10

                (0, 10)
                          .                      x
                              (2, −6)
                                      (3, −17)
                       10           −6 −17       f(x)
                        0            2 3         shape
                       IP           IP min
y
Step 4: Graph

     f(x) = x4 − 4x3 + 10

                (0, 10)
                          .                      x
                              (2, −6)
                                      (3, −17)
                       10           −6 −17       f(x)
                        0            2 3         shape
                       IP           IP min
y
Step 4: Graph

     f(x) = x4 − 4x3 + 10

                (0, 10)
                          .                      x
                              (2, −6)
                                      (3, −17)
                       10           −6 −17       f(x)
                        0            2 3         shape
                       IP           IP min
y
Step 4: Graph

     f(x) = x4 − 4x3 + 10

                (0, 10)
                          .                      x
                              (2, −6)
                                      (3, −17)
                       10           −6 −17       f(x)
                        0            2 3         shape
                       IP           IP min
y
Step 4: Graph

     f(x) = x4 − 4x3 + 10

                (0, 10)
                          .                      x
                              (2, −6)
                                      (3, −17)
                       10           −6 −17       f(x)
                        0            2 3         shape
                       IP           IP min
Outline
 Simple examples
    A cubic func on
    A quar c func on

 More Examples
   Points of nondifferen ability
   Horizontal asymptotes
   Ver cal asymptotes
   Trigonometric and polynomial together
   Logarithmic
Graphing a function with a cusp

 Example
                    √
 Graph f(x) = x +       |x|
Graphing a function with a cusp

 Example
                    √
 Graph f(x) = x +       |x|
 This func on looks strange because of the absolute value. But
 whenever we become nervous, we can just take cases.
Step 0: Finding Zeroes
              √
 f(x) = x +       |x|
     First, look at f by itself. We can tell that f(0) = 0 and that
     f(x) > 0 if x is posi ve.
Step 0: Finding Zeroes
              √
 f(x) = x +       |x|
     First, look at f by itself. We can tell that f(0) = 0 and that
     f(x) > 0 if x is posi ve.
     Are there nega ve numbers which are zeroes for f?
Step 0: Finding Zeroes
              √
 f(x) = x +       |x|
     First, look at f by itself. We can tell that f(0) = 0 and that
     f(x) > 0 if x is posi ve.
     Are there nega ve numbers which are zeroes for f?
                            √                 √
                       x + −x = 0 =⇒ −x = −x
                               −x = x2 =⇒ x2 + x = 0

     The only solu ons are x = 0 and x = −1.
Step 0: Asymptotic behavior
      √
 f(x) = x + |x|
       lim f(x) = ∞, because both terms tend to ∞.
     x→∞
Step 0: Asymptotic behavior
      √
 f(x) = x + |x|
       lim f(x) = ∞, because both terms tend to ∞.
     x→∞
      lim f(x) is indeterminate of the form −∞ + ∞. It’s the same
     x→−∞            √
     as lim (−y + y)
        y→+∞
Step 0: Asymptotic behavior
      √
 f(x) = x + |x|
       lim f(x) = ∞, because both terms tend to ∞.
     x→∞
      lim f(x) is indeterminate of the form −∞ + ∞. It’s the same
     x→−∞            √
     as lim (−y + y)
        y→+∞
                                                    √
                            √             √          y+y
                lim (−y +       y) = lim ( y − y) · √
               y→+∞                  y→∞             y+y
                                          y − y2
                                   = lim √       = −∞
                                     y→∞    y+y
Step 1: The derivative
                          √
 Remember, f(x) = x + |x|.
 To find f′ , first assume x > 0. Then
                              d (  √ )    1
                   f′ (x) =      x+ x =1+ √
                              dx         2 x
Step 1: The derivative
                          √
 Remember, f(x) = x + |x|.
 To find f′ , first assume x > 0. Then
                              d (  √ )    1
                   f′ (x) =      x+ x =1+ √
                              dx         2 x
 No ce
     f′ (x) > 0 when x > 0 (so no cri cal points here)
Step 1: The derivative
                          √
 Remember, f(x) = x + |x|.
 To find f′ , first assume x > 0. Then
                              d (  √ )    1
                   f′ (x) =      x+ x =1+ √
                              dx         2 x
 No ce
     f′ (x) > 0 when x > 0 (so no cri cal points here)
      lim+ f′ (x) = ∞ (so 0 is a cri cal point)
     x→0
Step 1: The derivative
                          √
 Remember, f(x) = x + |x|.
 To find f′ , first assume x > 0. Then
                              d (  √ )    1
                   f′ (x) =      x+ x =1+ √
                              dx         2 x
 No ce
     f′ (x) > 0 when x > 0 (so no cri cal points here)
      lim+ f′ (x) = ∞ (so 0 is a cri cal point)
     x→0
      lim f′ (x) = 1 (so the graph is asympto c to a line of slope 1)
     x→∞
Step 1: The derivative
            √
 Remember, f(x) = x + |x|.
 If x is nega ve, we have
                           d (    √ )        1
                  f′ (x) =     x + −x = 1 − √
                           dx              2 −x
 No ce
     lim− f′ (x) = −∞ (other side of the cri cal point)
     x→0
Step 1: The derivative
            √
 Remember, f(x) = x + |x|.
 If x is nega ve, we have
                           d (    √ )        1
                  f′ (x) =     x + −x = 1 − √
                           dx              2 −x
 No ce
     lim− f′ (x) = −∞ (other side of the cri cal point)
     x→0
       lim f′ (x) = 1 (asympto c to a line of slope 1)
     x→−∞
Step 1: The derivative
            √
 Remember, f(x) = x + |x|.
 If x is nega ve, we have
                           d (    √ )        1
                  f′ (x) =     x + −x = 1 − √
                           dx              2 −x
 No ce
     lim− f′ (x) = −∞ (other side of the cri cal point)
     x→0
       lim f′ (x) = 1 (asympto c to a line of slope 1)
     x→−∞
      ′
     f (x) = 0 when
          1        √     1         1          1
      1− √   = 0 =⇒ −x =   =⇒ −x =   =⇒ x = −
        2 −x             2         4          4
Step 1: Monotonicity
                                   1
                            1 + √
                                          if x > 0
                     ′
                    f (x) =        2 x
                            1 − √   1
                                          if x < 0
                                   2 −x
  We can’t make a mul -factor sign chart because of the absolute
 value, but we can test points in between cri cal points.

                                                     f′ (x)
                                 .
                                                     f(x)
Step 1: Monotonicity
                                   1
                            1 + √
                                          if x > 0
                     ′
                    f (x) =        2 x
                            1 − √   1
                                          if x < 0
                                   2 −x
  We can’t make a mul -factor sign chart because of the absolute
 value, but we can test points in between cri cal points.

                            0                        f′ (x)
                                 .
                           −1
                            4
                                                     f(x)
Step 1: Monotonicity
                                   1
                            1 + √
                                          if x > 0
                     ′
                    f (x) =        2 x
                            1 − √   1
                                          if x < 0
                                   2 −x
  We can’t make a mul -factor sign chart because of the absolute
 value, but we can test points in between cri cal points.

                            0    ∞                   f′ (x)
                                  .
                           −1
                            4
                                 0                   f(x)
Step 1: Monotonicity
                                   1
                            1 + √
                                          if x > 0
                     ′
                    f (x) =        2 x
                            1 − √   1
                                          if x < 0
                                   2 −x
  We can’t make a mul -factor sign chart because of the absolute
 value, but we can test points in between cri cal points.

                       +    0    ∞                   f′ (x)
                                  .
                           −1
                            4
                                 0                   f(x)
Step 1: Monotonicity
                                   1
                            1 + √
                                          if x > 0
                     ′
                    f (x) =        2 x
                            1 − √   1
                                          if x < 0
                                   2 −x
  We can’t make a mul -factor sign chart because of the absolute
 value, but we can test points in between cri cal points.

                       +   0− ∞                      f′ (x)
                               .
                           −4 0
                            1                        f(x)
Step 1: Monotonicity
                                   1
                            1 + √
                                          if x > 0
                     ′
                    f (x) =        2 x
                            1 − √   1
                                          if x < 0
                                   2 −x
  We can’t make a mul -factor sign chart because of the absolute
 value, but we can test points in between cri cal points.

                       +   0− ∞           +          f′ (x)
                               .
                           −4 0
                            1                        f(x)
Step 1: Monotonicity
                                   1
                            1 + √
                                          if x > 0
                     ′
                    f (x) =        2 x
                            1 − √   1
                                          if x < 0
                                   2 −x
  We can’t make a mul -factor sign chart because of the absolute
 value, but we can test points in between cri cal points.

                       + 0− ∞             +          f′ (x)
                             .
                       ↗ −4 0
                          1                          f(x)
Step 1: Monotonicity
                                   1
                            1 + √
                                          if x > 0
                     ′
                    f (x) =        2 x
                            1 − √   1
                                          if x < 0
                                   2 −x
  We can’t make a mul -factor sign chart because of the absolute
 value, but we can test points in between cri cal points.

                       + 0− ∞             +          f′ (x)
                              .
                       ↗ −4
                          1↘ 0                       f(x)
Step 1: Monotonicity
                                   1
                            1 + √
                                          if x > 0
                     ′
                    f (x) =        2 x
                            1 − √   1
                                          if x < 0
                                   2 −x
  We can’t make a mul -factor sign chart because of the absolute
 value, but we can test points in between cri cal points.

                       + 0− ∞             +          f′ (x)
                              .
                       ↗ −4
                          1↘ 0            ↗          f(x)
Step 1: Monotonicity
                                   1
                            1 + √
                                          if x > 0
                     ′
                    f (x) =        2 x
                            1 − √   1
                                          if x < 0
                                   2 −x
  We can’t make a mul -factor sign chart because of the absolute
 value, but we can test points in between cri cal points.

                       + 0− ∞             +          f′ (x)
                               .
                       ↗ −41↘ 0           ↗          f(x)
                         max
Step 1: Monotonicity
                                   1
                            1 + √
                                          if x > 0
                     ′
                    f (x) =        2 x
                            1 − √   1
                                          if x < 0
                                   2 −x
  We can’t make a mul -factor sign chart because of the absolute
 value, but we can test points in between cri cal points.

                       + 0− ∞             +          f′ (x)
                               .
                       ↗ −41↘ 0           ↗          f(x)
                         max min
Step 2: Concavity
              (                            )
                            d     1               1
   If x > 0, then f′′ (x) =    1 + x−1/2       = − x−3/2 This is
                            dx    2               4
   nega ve whenever x > 0.
Step 2: Concavity
              (                           )
                            d      1 −1/2       1
   If x > 0, then f′′ (x) =     1+ x        = − x−3/2 This is
                            dx     2            4
   nega ve whenever x > 0.     (              )
                   ′′       d      1     −1/2      1
   If x < 0, then f (x) =       1 − (−x)        = − (−x)−3/2
                            dx     2               4
   which is also always nega ve for nega ve x.
Step 2: Concavity
              (                            )
                            d       1 −1/2       1
   If x > 0, then f′′ (x) =      1+ x        = − x−3/2 This is
                            dx      2            4
   nega ve whenever x > 0.     (               )
                   ′′       d       1     −1/2      1
   If x < 0, then f (x) =        1 − (−x)        = − (−x)−3/2
                            dx      2               4
   which is also always nega ve for nega ve x.
                                1
   In other words, f′′ (x) = − |x|−3/2 .
                                4
Step 2: Concavity
              (                              )
                              d       1 −1/2       1
     If x > 0, then f′′ (x) =      1+ x        = − x−3/2 This is
                              dx      2            4
     nega ve whenever x > 0.     (               )
                     ′′       d       1     −1/2      1
     If x < 0, then f (x) =        1 − (−x)        = − (−x)−3/2
                              dx      2               4
     which is also always nega ve for nega ve x.
                                  1
     In other words, f′′ (x) = − |x|−3/2 .
                                  4
 Here is the sign chart:

                      −−       −∞        −−          f′′ (x)
                                 .
                      ⌢         0        ⌢           f(x)
Step 3: Synthesis
 Now we can put these things together.
                                    √
                         f(x) = x + |x|

    +1             + 0− ∞           +       f′
                                          +1 (x)
                            .
     ↗            ↗     1↘ 0       ↗       ↗monotonicity
    −∞            −− − −−
                        4 −∞       −−       f′′
                                          −∞ (x)
     ⌢            ⌢ 1 ⌢0           ⌢       ⌢concavity
    −∞ 0              4    0              +∞f(x)
        −1           −4 0
                        1                   shape
       zero          max min
Step 3: Synthesis
 Now we can put these things together.
                                    √
                         f(x) = x + |x|

    +1             + 0− ∞           +       f′
                                          +1 (x)
                            .
     ↗            ↗     1↘ 0       ↗       ↗monotonicity
    −∞            −− − −−
                        4 −∞       −−       f′′
                                          −∞ (x)
     ⌢            ⌢ 1 ⌢0           ⌢       ⌢concavity
    −∞ 0              4    0              +∞f(x)
        −1           −4 0
                        1                   shape
       zero          max min
Step 3: Synthesis
 Now we can put these things together.
                                    √
                         f(x) = x + |x|

    +1             + 0− ∞           +       f′
                                          +1 (x)
                            .
     ↗            ↗     1↘ 0       ↗       ↗monotonicity
    −∞            −− − −−
                        4 −∞       −−       f′′
                                          −∞ (x)
     ⌢            ⌢ 1 ⌢0           ⌢       ⌢concavity
    −∞ 0              4    0              +∞f(x)
        −1           −4 0
                        1                   shape
       zero          max min
Step 3: Synthesis
 Now we can put these things together.
                                    √
                         f(x) = x + |x|

    +1             + 0− ∞           +       f′
                                          +1 (x)
                            .
     ↗            ↗     1↘ 0       ↗       ↗monotonicity
    −∞            −− − −−
                        4 −∞       −−       f′′
                                          −∞ (x)
     ⌢            ⌢ 1 ⌢0           ⌢       ⌢concavity
    −∞ 0              4    0              +∞f(x)
        −1           −4 0
                        1                   shape
       zero          max min
Step 3: Synthesis
 Now we can put these things together.
                                    √
                         f(x) = x + |x|

    +1             + 0− ∞           +       f′
                                          +1 (x)
                            .
     ↗            ↗     1↘ 0       ↗       ↗monotonicity
    −∞            −− − −−
                        4 −∞       −−       f′′
                                          −∞ (x)
     ⌢            ⌢ 1 ⌢0           ⌢       ⌢concavity
    −∞ 0              4    0              +∞f(x)
        −1           −4 0
                        1                   shape
       zero          max min
Graph
                             √
                f(x) = x +       |x|



                       .                   x

                  1
        −∞0       4  0                 +∞ x
          −1    −1 0                      shape
                  4
         zero   max min
Graph
                               √
                  f(x) = x +       |x|


        (−1, 0)
                         .                   x

                    1
        −∞0         4  0                 +∞ x
          −1      −1 0                      shape
                    4
         zero     max min
Graph
                                   √
                  f(x) = x +           |x|


                  (− 1 , 1 )
                     4 4
        (−1, 0)
                               .                 x

                      1
        −∞0           4  0                   +∞ x
          −1        −1 0                        shape
                      4
         zero       max min
Graph
                                     √
                  f(x) = x +             |x|


                  (− 1 , 1 )
                     4 4
        (−1, 0)
                               .                   x
                                   (0, 0)
                      1
        −∞0           4  0                     +∞ x
          −1        −1 0                          shape
                      4
         zero       max min
Graph
                                     √
                  f(x) = x +             |x|


                  (− 1 , 1 )
                     4 4
        (−1, 0)
                               .                   x
                                   (0, 0)
                      1
        −∞0           4  0                     +∞ x
          −1        −1 0                          shape
                      4
         zero       max min
Graph
                                     √
                  f(x) = x +             |x|


                  (− 1 , 1 )
                     4 4
        (−1, 0)
                               .                   x
                                   (0, 0)
                      1
        −∞0           4  0                     +∞ x
          −1        −1 0                          shape
                      4
         zero       max min
Graph
                                     √
                  f(x) = x +             |x|


                  (− 1 , 1 )
                     4 4
        (−1, 0)
                               .                   x
                                   (0, 0)
                      1
        −∞0           4  0                     +∞ x
          −1        −1 0                          shape
                      4
         zero       max min
Example with Horizontal
Asymptotes

 Example
 Graph f(x) = xe−x
                 2
Example with Horizontal
Asymptotes

 Example
 Graph f(x) = xe−x
                   2




 Before taking deriva ves, we no ce that f is odd, that f(0) = 0, and
 lim f(x) = 0
 x→∞
Step 1: −xMonotonicity
                2
 If f(x) = xe       , then
                                           (       ) 2
            f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x
                            2      2

                     (     √ )(       √ ) −x2
                   = 1 − 2x 1 + 2x e

                                       . √0              √
                                                  1−         2x
                               0           1/2
                                                        √
                              √                   1 + 2x
                             − 0 1/2      0       f′ (x)
                              √          √
                             − 1/2         1/2    f(x)
Step 1: −xMonotonicity
                2
 If f(x) = xe       , then
                                           (       ) 2
            f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x
                            2      2

                     (     √ )(       √ ) −x2
                   = 1 − 2x 1 + 2x e

                +                      . √0              √
                                                  1−         2x
                               0           1/2
                                                        √
                              √                   1 + 2x
                             − 0 1/2      0       f′ (x)
                              √          √
                             − 1/2         1/2    f(x)
Step 1: −xMonotonicity
                2
 If f(x) = xe       , then
                                           (       ) 2
            f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x
                            2      2

                     (     √ )(       √ ) −x2
                   = 1 − 2x 1 + 2x e

                +                      + 0
                                        . √              √
                                                  1−         2x
                               0            1/2
                                                        √
                              √                   1 + 2x
                             − 0 1/2       0      f′ (x)
                              √           √
                             − 1/2          1/2   f(x)
Step 1: −xMonotonicity
                2
 If f(x) = xe       , then
                                           (       ) 2
            f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x
                            2      2

                     (     √ )(       √ ) −x2
                   = 1 − 2x 1 + 2x e

                +                      + 0
                                        . √       −          √
                                                      1−         2x
                               0            1/2
                                                            √
                              √                       1 + 2x
                             − 0 1/2       0          f′ (x)
                              √           √
                             − 1/2          1/2       f(x)
Step 1: −xMonotonicity
                2
 If f(x) = xe       , then
                                           (       ) 2
            f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x
                            2      2

                     (     √ )(       √ ) −x2
                   = 1 − 2x 1 + 2x e

                +                      + 0
                                        . √       −          √
                                                      1−         2x
                −              0            1/2
                                                            √
                              √                       1 + 2x
                             − 0 1/2       0          f′ (x)
                              √           √
                             − 1/2          1/2       f(x)
Step 1: −xMonotonicity
                2
 If f(x) = xe       , then
                                           (       ) 2
            f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x
                            2      2

                     (     √ )(       √ ) −x2
                   = 1 − 2x 1 + 2x e

                +                      + 0
                                        . √       −          √
                                                      1−         2x
                −              0       +    1/2
                                                            √
                              √                       1 + 2x
                             − 0 1/2       0          f′ (x)
                              √           √
                             − 1/2          1/2       f(x)
Step 1: −xMonotonicity
                2
 If f(x) = xe       , then
                                           (       ) 2
            f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x
                            2      2

                     (     √ )(       √ ) −x2
                   = 1 − 2x 1 + 2x e

                +                      + 0
                                        . √       −          √
                                                      1−         2x
                −              0       +    1/2
                                                  +         √
                              √                       1 + 2x
                             − 0 1/2       0          f′ (x)
                              √           √
                             − 1/2          1/2       f(x)
Step 1: −xMonotonicity
                2
 If f(x) = xe       , then
                                           (       ) 2
            f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x
                            2      2

                     (     √ )(       √ ) −x2
                   = 1 − 2x 1 + 2x e

                +                      + 0
                                        . √       −          √
                                                      1−         2x
                −              0       +    1/2
                                                  +         √
                              √                       1 + 2x
                −            − 0 1/2       0          f′ (x)
                              √           √
                             − 1/2          1/2       f(x)
Step 1: −xMonotonicity
                2
 If f(x) = xe       , then
                                           (       ) 2
            f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x
                            2      2

                     (     √ )(       √ ) −x2
                   = 1 − 2x 1 + 2x e

                +                  + 0
                                    . √       −          √
                                                  1−         2x
                −              0   +    1/2
                                              +         √
                               √                  1 + 2x
                −            − 0 1/2 + 0          f′ (x)
                               √      √
                             −   1/2    1/2       f(x)
Step 1: −xMonotonicity
                2
 If f(x) = xe       , then
                                           (       ) 2
            f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x
                            2      2

                     (     √ )(       √ ) −x2
                   = 1 − 2x 1 + 2x e

                +                  + 0
                                    . √       −          √
                                                  1−         2x
                −              0   +    1/2
                                              +         √
                               √                  1 + 2x
                −            − 0 1/2 + 0      −   f′ (x)
                               √      √
                ↘            −   1/2    1/2       f(x)
Step 1: −xMonotonicity
                2
 If f(x) = xe       , then
                                           (       ) 2
            f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x
                            2      2

                     (     √ )(       √ ) −x2
                   = 1 − 2x 1 + 2x e

                +                   + 0
                                     . √       −          √
                                                   1−         2x
                −              0    +    1/2
                                               +         √
                               √                   1 + 2x
                −            − 0 1/2 + 0       −   f′ (x)
                               √       √
                ↘            −   1/2 ↗   1/2       f(x)
Step 1: −xMonotonicity
                2
 If f(x) = xe       , then
                                           (       ) 2
            f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x
                            2      2

                     (     √ )(       √ ) −x2
                   = 1 − 2x 1 + 2x e

                +                   + 0
                                     . √       −          √
                                                   1−         2x
                −              0    +    1/2
                                               +         √
                               √                   1 + 2x
                −            − 0 1/2 + 0       −   f′ (x)
                               √       √
                ↘            −   1/2 ↗   1/2   ↘   f(x)
Step 1: −xMonotonicity
                2
 If f(x) = xe       , then
                                           (       ) 2
            f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x
                            2      2

                     (     √ )(       √ ) −x2
                   = 1 − 2x 1 + 2x e

                +                   + 0
                                     . √       −          √
                                                   1−         2x
                −              0    +    1/2
                                               +         √
                               √                   1 + 2x
                −            − 0 1/2 + 0       −   f′ (x)
                               √       √
                ↘            −   1/2 ↗   1/2   ↘   f(x)
Step 1: −xMonotonicity
                2
 If f(x) = xe       , then
                                           (       ) 2
            f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x
                            2      2

                     (     √ )(       √ ) −x2
                   = 1 − 2x 1 + 2x e

                +                   + 0
                                     . √       −          √
                                                   1−         2x
                −              0    +    1/2
                                               +         √
                               √                   1 + 2x
                −            − 0 1/2 + 0       −   f′ (x)
                               √       √
                ↘            −   1/2 ↗   1/2   ↘   f(x)
                              min
Step 1: −xMonotonicity
                2
 If f(x) = xe       , then
                                           (       ) 2
            f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x
                            2      2

                     (     √ )(       √ ) −x2
                   = 1 − 2x 1 + 2x e

                +                   + 0
                                     . √       −          √
                                                   1−         2x
                −              0    +    1/2
                                               +         √
                               √                   1 + 2x
                −            − 0 1/2 + 0       −   f′ (x)
                               √       √
                ↘            −   1/2 ↗   1/2   ↘   f(x)
                              min      max
Step 2: Concavity
   ′     2 −x            2
 If f (x) = (1 − 2x )e       , we know
                                               (        ) 2
     f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x
                       2               2



           = 2x(2x2 − 3)e−x
                                 2




                                0.              2x
                                0
                                          0     √       √
                                         √          2x − 3
                   0                      3/2
                                                √       √
                  √                                 2x + 3
                 − 0 3/2        0         0     f′′ (x)
                  √                      √
                 − 3/2          0         3/2   f(x)
Step 2: Concavity
   ′     2 −x            2
 If f (x) = (1 − 2x )e       , we know
                                               (        ) 2
     f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x
                       2               2



           = 2x(2x2 − 3)e−x
                                 2




             −                  0.              2x
                                0
                                          0     √       √
                                         √          2x − 3
                   0                      3/2
                                                √       √
                  √                                 2x + 3
                 − 0 3/2        0         0     f′′ (x)
                  √                      √
                 − 3/2          0         3/2   f(x)
Step 2: Concavity
   ′     2 −x            2
 If f (x) = (1 − 2x )e       , we know
                                               (        ) 2
     f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x
                       2               2



           = 2x(2x2 − 3)e−x
                                 2




             −             −    0.              2x
                                0
                                          0     √       √
                                         √          2x − 3
                   0                      3/2
                                                √       √
                  √                                 2x + 3
                 − 0 3/2        0         0     f′′ (x)
                  √                      √
                 − 3/2          0         3/2   f(x)
Step 2: Concavity
   ′     2 −x            2
 If f (x) = (1 − 2x )e       , we know
                                               (        ) 2
     f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x
                       2               2



           = 2x(2x2 − 3)e−x
                                 2




             −             −    0.   +
                                                2x
                                0
                                          0     √       √
                                         √          2x − 3
                   0                      3/2
                                                √       √
                  √                                 2x + 3
                 − 0 3/2        0         0     f′′ (x)
                  √                      √
                 − 3/2          0         3/2   f(x)
Step 2: Concavity
   ′     2 −x            2
 If f (x) = (1 − 2x )e       , we know
                                               (        ) 2
     f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x
                       2               2



           = 2x(2x2 − 3)e−x
                                 2




             −             −    0.   +          +
                                                    2x
                                0
                                          0         √       √
                                         √              2x − 3
                   0                      3/2
                                                    √       √
                  √                                     2x + 3
                 − 0 3/2        0         0         f′′ (x)
                  √                      √
                 − 3/2          0         3/2       f(x)
Step 2: Concavity
   ′     2 −x            2
 If f (x) = (1 − 2x )e       , we know
                                               (        ) 2
     f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x
                       2               2



           = 2x(2x2 − 3)e−x
                                 2




             −             −    0.   +          +
                                                    2x
                                0
             −                            0         √       √
                                         √              2x − 3
                   0                      3/2
                                                    √       √
                  √                                     2x + 3
                 − 0 3/2        0         0         f′′ (x)
                  √                      √
                 − 3/2          0         3/2       f(x)
Step 2: Concavity
   ′     2 −x            2
 If f (x) = (1 − 2x )e       , we know
                                               (        ) 2
     f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x
                       2               2



           = 2x(2x2 − 3)e−x
                                 2




             −             −    0.   +          +
                                                    2x
                                0
             −             −              0         √       √
                                         √              2x − 3
                   0                      3/2
                                                    √       √
                  √                                     2x + 3
                 − 0 3/2        0         0         f′′ (x)
                  √                      √
                 − 3/2          0         3/2       f(x)
Step 2: Concavity
   ′     2 −x            2
 If f (x) = (1 − 2x )e       , we know
                                               (        ) 2
     f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x
                       2               2



           = 2x(2x2 − 3)e−x
                                 2




             −             −    0.   +          +
                                                    2x
                                0
             −             −         −    0         √       √
                                         √              2x − 3
                   0                      3/2
                                                    √       √
                  √                                     2x + 3
                 − 0 3/2        0         0         f′′ (x)
                  √                      √
                 − 3/2          0         3/2       f(x)
Step 2: Concavity
   ′     2 −x            2
 If f (x) = (1 − 2x )e       , we know
                                               (        ) 2
     f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x
                       2               2



           = 2x(2x2 − 3)e−x
                                 2




             −             −    0.   +          +
                                                    2x
                                0
             −             −         −    0   +     √       √
                                         √              2x − 3
                   0                      3/2
                                                    √       √
                  √                                     2x + 3
                 − 0 3/2        0         0         f′′ (x)
                  √                      √
                 − 3/2          0         3/2       f(x)
Step 2: Concavity
   ′     2 −x            2
 If f (x) = (1 − 2x )e       , we know
                                               (        ) 2
     f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x
                       2               2



           = 2x(2x2 − 3)e−x
                                 2




             −             −    0.   +          +
                                                    2x
                                0
             −             −         −    0   +     √       √
                                         √              2x − 3
             −     0                      3/2
                                                    √       √
                  √                                     2x + 3
                 − 0 3/2        0         0         f′′ (x)
                  √                      √
                 − 3/2          0         3/2       f(x)
Step 2: Concavity
   ′     2 −x            2
 If f (x) = (1 − 2x )e       , we know
                                               (        ) 2
     f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x
                       2               2



           = 2x(2x2 − 3)e−x
                                 2




             −           −      0.   +          +
                                                    2x
                                0
             −           −           −    0   +     √       √
                                         √              2x − 3
             −     0 +                    3/2
                                                    √       √
                  √                                     2x + 3
                 − 0 3/2        0         0         f′′ (x)
                  √                      √
                 − 3/2          0         3/2       f(x)
Step 2: Concavity
   ′     2 −x            2
 If f (x) = (1 − 2x )e       , we know
                                               (        ) 2
     f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x
                       2               2



           = 2x(2x2 − 3)e−x
                                 2




             −           −      0.   +          +
                                                    2x
                                0
             −           −           − 0   +        √       √
                                      √                 2x − 3
             −     0 +               + 3/2          √       √
                  √                                     2x + 3
                 − 0 3/2        0         0         f′′ (x)
                  √                      √
                 − 3/2          0         3/2       f(x)
Step 2: Concavity
   ′     2 −x            2
 If f (x) = (1 − 2x )e       , we know
                                               (        ) 2
     f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x
                       2               2



           = 2x(2x2 − 3)e−x
                                 2




             −           −      0.   +          +
                                                    2x
                                0
             −           −           − 0   +        √       √
                                      √                 2x − 3
             −     0 +               + 3/2 +        √       √
                  √                                     2x + 3
                 − 0 3/2        0         0         f′′ (x)
                  √                      √
                 − 3/2          0         3/2       f(x)
Step 2: Concavity
   ′     2 −x            2
 If f (x) = (1 − 2x )e       , we know
                                               (        ) 2
     f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x
                       2               2



           = 2x(2x2 − 3)e−x
                                 2




             −           −      0.   +          +
                                                    2x
                                0
             −           −           − 0   +        √       √
                                      √                 2x − 3
             −   0 +                 + 3/2 +        √       √
                √                                       2x + 3
            −− − 0 3/2          0         0         f′′ (x)
                √                        √
               − 3/2            0         3/2       f(x)
Step 2: Concavity
   ′     2 −x            2
 If f (x) = (1 − 2x )e       , we know
                                               (        ) 2
     f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x
                       2               2



           = 2x(2x2 − 3)e−x
                                 2




             −           −      0.   +          +
                                                    2x
                                0
             −           −           − 0   +        √       √
                                      √                 2x − 3
             −   0 +                 + 3/2 +        √       √
                √                                       2x + 3
            −− − 0 3/2
                     ++ 0                 0         f′′ (x)
                √                        √
               − 3/2    0                 3/2       f(x)
Step 2: Concavity
   ′     2 −x            2
 If f (x) = (1 − 2x )e       , we know
                                               (        ) 2
     f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x
                       2               2



           = 2x(2x2 − 3)e−x
                                 2




             −           −      0.   +    +
                                                2x
                                0
             −           −           − 0   +    √       √
                                      √             2x − 3
             −   0 +                 + 3/2 +    √       √
                √                                   2x + 3
            −− − 0 3/2
                     ++ 0 −− 0                  f′′ (x)
                √           √
               − 3/2    0    3/2                f(x)
Step 2: Concavity
   ′     2 −x            2
 If f (x) = (1 − 2x )e       , we know
                                               (        ) 2
     f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x
                       2               2



           = 2x(2x2 − 3)e−x
                                 2




             −           −      0.   +    +
                                                2x
                                0
             −           −           − 0   +    √       √
                                      √             2x − 3
             −   0 +                 + 3/2 +    √       √
                √                                   2x + 3
            −− − 0 3/2
                     ++ 0 −− 0 ++               f′′ (x)
            ⌢   √           √
               − 3/2    0    3/2                f(x)
Step 2: Concavity
   ′     2 −x            2
 If f (x) = (1 − 2x )e       , we know
                                               (        ) 2
     f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x
                       2               2



           = 2x(2x2 − 3)e−x
                                 2




             −           −      0.   +    +
                                                2x
                                0
             −           −           − 0   +    √       √
                                      √             2x − 3
             −   0 +                 + 3/2 +    √       √
                √                                   2x + 3
            −− − 0 3/2
                     ++ 0 −− 0 ++               f′′ (x)
            ⌢   √ ⌣         √
               − 3/2    0    3/2                f(x)
Step 2: Concavity
   ′     2 −x            2
 If f (x) = (1 − 2x )e       , we know
                                               (        ) 2
     f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x
                       2               2



           = 2x(2x2 − 3)e−x
                                 2




             −           −      0.   +    +
                                                2x
                                0
             −           −           − 0   +    √       √
                                      √             2x − 3
             −   0 +                 + 3/2 +    √       √
                √                                   2x + 3
            −− − 0 3/2
                     ++ 0 −− 0 ++               f′′ (x)
            ⌢   √ ⌣         √
               − 3/2    0 ⌢ 3/2                 f(x)
Step 2: Concavity
   ′     2 −x            2
 If f (x) = (1 − 2x )e       , we know
                                               (        ) 2
     f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x
                       2               2



           = 2x(2x2 − 3)e−x
                                 2




             −           −      0.   +    +
                                                2x
                                0
             −           −           − 0   +    √       √
                                      √             2x − 3
             −   0 +                 + 3/2 +    √       √
                √                                   2x + 3
            −− − 0 3/2
                     ++ 0 −− 0 ++               f′′ (x)
            ⌢   √ ⌣         √
               − 3/2    0 ⌢ 3/2 ⌣               f(x)
Step 2: Concavity
   ′     2 −x            2
 If f (x) = (1 − 2x )e       , we know
                                               (        ) 2
     f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x
                       2               2



           = 2x(2x2 − 3)e−x
                                 2




             −           −      0.   +    +
                                                2x
                                0
             −           −           − 0   +    √       √
                                      √             2x − 3
             −   0 +                 + 3/2 +    √       √
                √                                   2x + 3
            −− − 0 3/2
                     ++ 0 −− 0 ++               f′′ (x)
            ⌢   √ ⌣         √
               − 3/2    0 ⌢ 3/2 ⌣               f(x)
Step 2: Concavity
   ′     2 −x            2
 If f (x) = (1 − 2x )e       , we know
                                               (        ) 2
     f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x
                       2               2



           = 2x(2x2 − 3)e−x
                                 2




             −           −      0.   +    +
                                                2x
                                0
             −           −           − 0   +    √       √
                                      √             2x − 3
             −    0 +                + 3/2 +    √       √
                √                                   2x + 3
            −− − 0 3/2
                     ++ 0 −− 0 ++               f′′ (x)
            ⌢   √ ⌣         √
               − 3/2    0 ⌢ 3/2 ⌣               f(x)
                 IP
Step 2: Concavity
   ′     2 −x            2
 If f (x) = (1 − 2x )e       , we know
                                               (        ) 2
     f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x
                       2               2



           = 2x(2x2 − 3)e−x
                                 2




             −           −      0.   +    +
                                                2x
                                0
             −           −           − 0   +    √       √
                                      √             2x − 3
             −    0 +                + 3/2 +    √       √
                √                                   2x + 3
            −− − 0 3/2
                     ++ 0 −− 0 ++               f′′ (x)
            ⌢   √ ⌣         √
               − 3/2     0 ⌢ 3/2 ⌣              f(x)
                 IP     IP
Step 2: Concavity
   ′     2 −x            2
 If f (x) = (1 − 2x )e       , we know
                                               (        ) 2
     f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x
                       2               2



           = 2x(2x2 − 3)e−x
                                 2




             −           −      0.   +    +
                                                2x
                                0
             −           −           − 0   +    √       √
                                      √             2x − 3
             −    0 +                + 3/2 +    √       √
                √                                   2x + 3
            −− − 0 3/2
                     ++ 0 −− 0 ++               f′′ (x)
            ⌢   √ ⌣         √
               − 3/2     0 ⌢ 3/2 ⌣              f(x)
                 IP     IP   IP
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
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Lesson 21: Curve Sketching (slides)

  • 1. Sec on 4.4 Curve Sketching V63.0121.001: Calculus I Professor Ma hew Leingang New York University April 13, 2011 .
  • 2. Announcements Quiz 4 on Sec ons 3.3, 3.4, 3.5, and 3.7 this week (April 14/15) Quiz 5 on Sec ons 4.1–4.4 April 28/29 Final Exam Thursday May 12, 2:00–3:50pm I am teaching Calc II MW 2:00pm and Calc III TR 2:00pm both Fall ’11 and Spring ’12
  • 3. Objectives given a func on, graph it completely, indica ng zeroes (if easy) asymptotes if applicable cri cal points local/global max/min inflec on points
  • 4. Why? Graphing func ons is like dissec on
  • 5. Why? Graphing func ons is like dissec on … or diagramming sentences
  • 6. Why? Graphing func ons is like dissec on … or diagramming sentences You can really know a lot about a func on when you know all of its anatomy.
  • 7. The Increasing/Decreasing Test Theorem (The Increasing/Decreasing Test) If f′ > 0 on (a, b), then f is increasing on (a, b). If f′ < 0 on (a, b), then f is decreasing on (a, b). Example f(x) f(x) = x3 + x2 .
  • 8. The Increasing/Decreasing Test Theorem (The Increasing/Decreasing Test) If f′ > 0 on (a, b), then f is increasing on (a, b). If f′ < 0 on (a, b), then f is decreasing on (a, b). Example f(x) f′ (x) f(x) = x3 + x2 f′ (x) = 3x2 + 2x .
  • 9. Testing for Concavity Theorem (Concavity Test) If f′′ (x) > 0 for all x in (a, b), then the graph of f is concave upward on (a, b) If f′′ (x) < 0 for all x in (a, b), then the graph of f is concave downward on (a, b). Example f(x) f(x) = x3 + x2 .
  • 10. Testing for Concavity Theorem (Concavity Test) If f′′ (x) > 0 for all x in (a, b), then the graph of f is concave upward on (a, b) If f′′ (x) < 0 for all x in (a, b), then the graph of f is concave downward on (a, b). Example f′ (x) f(x) f(x) = x3 + x2 f′ (x) = 3x2 + 2x .
  • 11. Testing for Concavity Theorem (Concavity Test) If f′′ (x) > 0 for all x in (a, b), then the graph of f is concave upward on (a, b) If f′′ (x) < 0 for all x in (a, b), then the graph of f is concave downward on (a, b). Example f′′ (x) f′ (x) f(x) f(x) = x3 + x2 f′ (x) = 3x2 + 2x . f′′ (x) = 6x + 2
  • 12. Graphing Checklist To graph a func on f, follow this plan: 0. Find when f is posi ve, nega ve, zero, not defined.
  • 13. Graphing Checklist To graph a func on f, follow this plan: 0. Find when f is posi ve, nega ve, zero, not defined. 1. Find f′ and form its sign chart. Conclude informa on about increasing/decreasing and local max/min.
  • 14. Graphing Checklist To graph a func on f, follow this plan: 0. Find when f is posi ve, nega ve, zero, not defined. 1. Find f′ and form its sign chart. Conclude informa on about increasing/decreasing and local max/min. 2. Find f′′ and form its sign chart. Conclude concave up/concave down and inflec on.
  • 15. Graphing Checklist To graph a func on f, follow this plan: 3. Put together a big chart to assemble monotonicity and concavity data
  • 16. Graphing Checklist To graph a func on f, follow this plan: 3. Put together a big chart to assemble monotonicity and concavity data 4. Graph!
  • 17. Outline Simple examples A cubic func on A quar c func on More Examples Points of nondifferen ability Horizontal asymptotes Ver cal asymptotes Trigonometric and polynomial together Logarithmic
  • 18. Graphing a cubic Example Graph f(x) = 2x3 − 3x2 − 12x.
  • 19. Graphing a cubic Example Graph f(x) = 2x3 − 3x2 − 12x. (Step 0) First, let’s find the zeros. We can at least factor out one power of x: f(x) = x(2x2 − 3x − 12) so f(0) = 0. The other factor is a quadra c, so we the other two roots are √ √ 3 ± 32 − 4(2)(−12) 3 ± 105 x= = 4 4 It’s OK to skip this step for now since the roots are so complicated.
  • 20. Step 1: Monotonicity f(x) = 2x3 − 3x2 − 12x =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: .
  • 21. Step 1: Monotonicity f(x) = 2x3 − 3x2 − 12x =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: . x−2 2 x+1 −1 f′ (x) −1 2 f(x)
  • 22. Step 1: Monotonicity f(x) = 2x3 − 3x2 − 12x =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: − .− + x−2 2 x+1 −1 f′ (x) −1 2 f(x)
  • 23. Step 1: Monotonicity f(x) = 2x3 − 3x2 − 12x =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: − .− + x−2 2 − + + x+1 −1 f′ (x) −1 2 f(x)
  • 24. Step 1: Monotonicity f(x) = 2x3 − 3x2 − 12x =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: − .− + x−2 2 − + + x+1 −1 f′ (x) + −1 2 f(x)
  • 25. Step 1: Monotonicity f(x) = 2x3 − 3x2 − 12x =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: − .− + x−2 2 − + + x+1 −1 f′ (x) + − −1 2 f(x)
  • 26. Step 1: Monotonicity f(x) = 2x3 − 3x2 − 12x =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: − .− + x−2 2 − + + x+1 −1 f′ (x) + − + −1 2 f(x)
  • 27. Step 1: Monotonicity f(x) = 2x3 − 3x2 − 12x =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: − .− + x−2 2 − + + x+1 −1 f′ (x) + − + ↗−1 2 f(x)
  • 28. Step 1: Monotonicity f(x) = 2x3 − 3x2 − 12x =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: − .− + x−2 2 − + + x+1 −1 f′ (x) + − + ↗−1 ↘ 2 f(x)
  • 29. Step 1: Monotonicity f(x) = 2x3 − 3x2 − 12x =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: − .− + x−2 2 − + + x+1 −1 f′ (x) + − + ↗−1 ↘ 2 ↗ f(x)
  • 30. Step 1: Monotonicity f(x) = 2x3 − 3x2 − 12x =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: − .− + x−2 2 − + + x+1 −1 f′ (x) + − + ↗−1 ↘ 2 ↗ f(x) max
  • 31. Step 1: Monotonicity f(x) = 2x3 − 3x2 − 12x =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: − .− + x−2 2 − + + x+1 −1 f′ (x) + − + ↗−1 ↘ 2 ↗ f(x) max min
  • 32. Step 2: Concavity f′ (x) = 6x2 − 6x − 12 =⇒ f′′ (x) = 12x − 6 = 6(2x − 1) Another sign chart: .
  • 33. Step 2: Concavity f′ (x) = 6x2 − 6x − 12 =⇒ f′′ (x) = 12x − 6 = 6(2x − 1) Another sign chart: . f′′ (x) 1/2 f(x)
  • 34. Step 2: Concavity f′ (x) = 6x2 − 6x − 12 =⇒ f′′ (x) = 12x − 6 = 6(2x − 1) Another sign chart: . −− f′′ (x) 1/2 f(x)
  • 35. Step 2: Concavity f′ (x) = 6x2 − 6x − 12 =⇒ f′′ (x) = 12x − 6 = 6(2x − 1) Another sign chart: . −− ++ f′′ (x) 1/2 f(x)
  • 36. Step 2: Concavity f′ (x) = 6x2 − 6x − 12 =⇒ f′′ (x) = 12x − 6 = 6(2x − 1) Another sign chart: . −− ++ f′′ (x) ⌢ 1/2 f(x)
  • 37. Step 2: Concavity f′ (x) = 6x2 − 6x − 12 =⇒ f′′ (x) = 12x − 6 = 6(2x − 1) Another sign chart: . −− ++ f′′ (x) ⌢ 1/2 ⌣ f(x)
  • 38. Step 2: Concavity f′ (x) = 6x2 − 6x − 12 =⇒ f′′ (x) = 12x − 6 = 6(2x − 1) Another sign chart: . −− ++ f′′ (x) ⌢ 1/2 ⌣ f(x) IP
  • 39. Step 3: One sign chart to rule them all Remember, f(x) = 2x3 − 3x2 − 12x. .
  • 40. Step 3: One sign chart to rule them all Remember, f(x) = 2x3 − 3x2 − 12x. + −. − + f′ (x) ↗−1 ↘ ↘ 2 ↗ monotonicity
  • 41. Step 3: One sign chart to rule them all Remember, f(x) = 2x3 − 3x2 − 12x. + −. − + f′ (x) ↗−1 ↘ ↘ 2 ↗ monotonicity −− −− ++ ++ f′′ (x) ⌢ ⌢ 1/2 ⌣ ⌣ concavity
  • 42. Step 3: One sign chart to rule them all Remember, f(x) = 2x3 − 3x2 − 12x. + −. − + f′ (x) ↗−1 ↘ ↘ 2 ↗ monotonicity −− −− ++ ++ f′′ (x) ⌢ ⌢ 1/2 ⌣ ⌣ concavity 7 −61/2 −20 f(x) −1 1/2 2 shape of f max IP min
  • 43. monotonicity and concavity II I . III IV
  • 44. monotonicity and concavity decreasing, concave down II I . III IV
  • 45. monotonicity and concavity increasing, decreasing, concave concave down down II I . III IV
  • 46. monotonicity and concavity increasing, decreasing, concave concave down down II I . III IV decreasing, concave up
  • 47. monotonicity and concavity increasing, decreasing, concave concave down down II I . III IV decreasing, increasing, concave concave up up
  • 48. Step 3: One sign chart to rule them all Remember, f(x) = 2x3 − 3x2 − 12x. + −. − + f′ (x) ↗−1 ↘ ↘ 2 ↗ monotonicity −− −− ++ ++ f′′ (x) ⌢ ⌢ 1/2 ⌣ ⌣ concavity 7 −61/2 −20 f(x) −1 1/2 2 shape of f max IP min
  • 49. Step 3: One sign chart to rule them all Remember, f(x) = 2x3 − 3x2 − 12x. + −. − + f′ (x) ↗−1 ↘ ↘ 2 ↗ monotonicity −− −− ++ ++ f′′ (x) ⌢ ⌢ 1/2 ⌣ ⌣ concavity 7 −61/2 −20 f(x) −1 1/2 2 shape of f max IP min
  • 50. Step 3: One sign chart to rule them all Remember, f(x) = 2x3 − 3x2 − 12x. + −. − + f′ (x) ↗−1 ↘ ↘ 2 ↗ monotonicity −− −− ++ ++ f′′ (x) ⌢ ⌢ 1/2 ⌣ ⌣ concavity 7 −61/2 −20 f(x) −1 1/2 2 shape of f max IP min
  • 51. Step 3: One sign chart to rule them all Remember, f(x) = 2x3 − 3x2 − 12x. + −. − + f′ (x) ↗−1 ↘ ↘ 2 ↗ monotonicity −− −− ++ ++ f′′ (x) ⌢ ⌢ 1/2 ⌣ ⌣ concavity 7 −61/2 −20 f(x) −1 1/2 2 shape of f max IP min
  • 52. f(x) Step 4: Graph f(x) = 2x3 − 3x2 − 12x ( √ ) (−1, 7) 3− 105 4 ,0 (0, 0) . ( x√ ) (1/2, −61/2) 3+ 105 4 ,0 (2, −20) 7 −61/2 −20 f(x) −1 1/2 2 shape of f max IP min
  • 53. f(x) Step 4: Graph f(x) = 2x3 − 3x2 − 12x ( √ ) (−1, 7) 3− 105 4 ,0 (0, 0) . ( x√ ) (1/2, −61/2) 3+ 105 4 ,0 (2, −20) 7 −61/2 −20 f(x) −1 1/2 2 shape of f max IP min
  • 54. f(x) Step 4: Graph f(x) = 2x3 − 3x2 − 12x ( √ ) (−1, 7) 3− 105 4 ,0 (0, 0) . ( x√ ) (1/2, −61/2) 3+ 105 4 ,0 (2, −20) 7 −61/2 −20 f(x) −1 1/2 2 shape of f max IP min
  • 55. f(x) Step 4: Graph f(x) = 2x3 − 3x2 − 12x ( √ ) (−1, 7) 3− 105 4 ,0 (0, 0) . ( x√ ) (1/2, −61/2) 3+ 105 4 ,0 (2, −20) 7 −61/2 −20 f(x) −1 1/2 2 shape of f max IP min
  • 56. f(x) Step 4: Graph f(x) = 2x3 − 3x2 − 12x ( √ ) (−1, 7) 3− 105 4 ,0 (0, 0) . ( x√ ) (1/2, −61/2) 3+ 105 4 ,0 (2, −20) 7 −61/2 −20 f(x) −1 1/2 2 shape of f max IP min
  • 57. Graphing a quartic Example Graph f(x) = x4 − 4x3 + 10
  • 58. Graphing a quartic Example Graph f(x) = x4 − 4x3 + 10 (Step 0) We know f(0) = 10 and lim f(x) = +∞. Not too many x→±∞ other points on the graph are evident.
  • 59. Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3)
  • 60. Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. .
  • 61. Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. 0. 4x2 0
  • 62. Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. +0. 4x2 0
  • 63. Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. +0. + 4x2 0
  • 64. Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. +0. + + 4x2 0
  • 65. Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. +0. + + 4x2 0 0 (x − 3) 3
  • 66. Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. +0 . + + 4x2 0 − 0 (x − 3) 3
  • 67. Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. +0 . + + 4x2 0 − − 0 (x − 3) 3
  • 68. Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. +0 . + + 4x2 0 − − 0+ (x − 3) 3
  • 69. Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. +0 . + + 4x2 0 − − 0+ (x − 3) 3 f′ (x) 0 0 0 3 f(x)
  • 70. Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. +0 . + + 4x2 0 − − 0+ (x − 3) 3 f′ (x) −0 0 0 3 f(x)
  • 71. Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. +0 . + + 4x2 0 − − 0+ (x − 3) 3 f′ (x) −0 − 0 0 3 f(x)
  • 72. Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. +0 . + + 4x2 0 − − 0+ (x − 3) 3 ′ −0 − 0 + f (x) 0 3 f(x)
  • 73. Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. +0 . + + 4x2 0 − − 0+ (x − 3) 3 ′ −0 − 0 + f (x) ↘0 3 f(x)
  • 74. Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. +0 . + + 4x2 0 − − 0+ (x − 3) 3 ′ −0 − 0 + f (x) ↘0 ↘ 3 f(x)
  • 75. Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. +0 . + + 4x2 0 − − 0+ (x − 3) 3 ′ −0 − 0 + f (x) ↘0 ↘ 3 ↗ f(x)
  • 76. Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. +0 . + + 4x2 0 − − 0+ (x − 3) 3 ′ −0 − 0 + f (x) ↘0 ↘ 3 ↗ f(x) min
  • 77. Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) .
  • 78. Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: .
  • 79. Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: 0 . 12x 0
  • 80. Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: −0 . 12x 0
  • 81. Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: −0 . + 12x 0
  • 82. Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: −0 . + + 12x 0
  • 83. Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: −0 . + + 12x 0 0 x−2 2
  • 84. Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: −0. + + 12x 0 − 0 x−2 2
  • 85. Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: −0. + + 12x 0 − − 0 x−2 2
  • 86. Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: −0. + + 12x 0 − − 0 + x−2 2
  • 87. Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: −0. + + 12x 0 − − 0 + x−2 2 f′′ (x) 0 0 0 2 f(x)
  • 88. Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: −0. + + 12x 0 − − 0 + x−2 2 f′′ (x) ++0 0 0 2 f(x)
  • 89. Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: −0. + + 12x 0 − − 0 + x−2 2 f′′ (x) ++0 −− 0 0 2 f(x)
  • 90. Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: −0. + + 12x 0 − − 0 + x−2 2 f′′ (x) ++0 −− 0 ++ 0 2 f(x)
  • 91. Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: −0. + + 12x 0 − − 0 + x−2 2 f′′ (x) ++0 −− 0 ++ ⌣0 2 f(x)
  • 92. Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: −0. + + 12x 0 − − 0 + x−2 2 f′′ (x) ++0 −− 0 ++ ⌣0 ⌢ 2 f(x)
  • 93. Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: −0. + + 12x 0 − − 0 + x−2 2 f′′ (x) ++0 −− 0 ++ ⌣0 ⌢ 2 ⌣ f(x)
  • 94. Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: −0. + + 12x 0 − − 0 + x−2 2 f′′ (x) ++0 −− 0 ++ ⌣0 ⌢ 2 ⌣ f(x) IP
  • 95. Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: −0. + + 12x 0 − −0 + x−2 2 f′′ (x) ++0 −− 0 ++ ⌣0 ⌢ 2 ⌣ f(x) IP IP
  • 96. Step 3: Grand Unified Sign Chart . Remember, f(x) = x − 4x3 + 10. 4 −0 − −0+ f′ (x) ↘0 ↘ ↘3↗ monotonicity f′′ (x) ++0 −− 0++ ++ ⌣0 ⌢ 2⌣ ⌣ concavity 10 −6 −17 f(x) 0 2 3 shape IP IP min
  • 97. Step 3: Grand Unified Sign Chart . Remember, f(x) = x − 4x3 + 10. 4 −0 − −0+ f′ (x) ↘0 ↘ ↘3↗ monotonicity f′′ (x) ++0 −− 0++ ++ ⌣0 ⌢ 2⌣ ⌣ concavity 10 −6 −17 f(x) 0 2 3 shape IP IP min
  • 98. Step 3: Grand Unified Sign Chart . Remember, f(x) = x − 4x3 + 10. 4 −0 − −0+ f′ (x) ↘0 ↘ ↘3↗ monotonicity f′′ (x) ++0 −− 0++ ++ ⌣0 ⌢ 2⌣ ⌣ concavity 10 −6 −17 f(x) 0 2 3 shape IP IP min
  • 99. Step 3: Grand Unified Sign Chart . Remember, f(x) = x − 4x3 + 10. 4 −0 − −0+ f′ (x) ↘0 ↘ ↘3↗ monotonicity f′′ (x) ++0 −− 0++ ++ ⌣0 ⌢ 2⌣ ⌣ concavity 10 −6 −17 f(x) 0 2 3 shape IP IP min
  • 100. Step 3: Grand Unified Sign Chart . Remember, f(x) = x − 4x3 + 10. 4 −0 − −0+ f′ (x) ↘0 ↘ ↘3↗ monotonicity f′′ (x) ++0 −− 0++ ++ ⌣0 ⌢ 2⌣ ⌣ concavity 10 −6 −17 f(x) 0 2 3 shape IP IP min
  • 101. y Step 4: Graph f(x) = x4 − 4x3 + 10 (0, 10) . x (2, −6) (3, −17) 10 −6 −17 f(x) 0 2 3 shape IP IP min
  • 102. y Step 4: Graph f(x) = x4 − 4x3 + 10 (0, 10) . x (2, −6) (3, −17) 10 −6 −17 f(x) 0 2 3 shape IP IP min
  • 103. y Step 4: Graph f(x) = x4 − 4x3 + 10 (0, 10) . x (2, −6) (3, −17) 10 −6 −17 f(x) 0 2 3 shape IP IP min
  • 104. y Step 4: Graph f(x) = x4 − 4x3 + 10 (0, 10) . x (2, −6) (3, −17) 10 −6 −17 f(x) 0 2 3 shape IP IP min
  • 105. y Step 4: Graph f(x) = x4 − 4x3 + 10 (0, 10) . x (2, −6) (3, −17) 10 −6 −17 f(x) 0 2 3 shape IP IP min
  • 106. Outline Simple examples A cubic func on A quar c func on More Examples Points of nondifferen ability Horizontal asymptotes Ver cal asymptotes Trigonometric and polynomial together Logarithmic
  • 107. Graphing a function with a cusp Example √ Graph f(x) = x + |x|
  • 108. Graphing a function with a cusp Example √ Graph f(x) = x + |x| This func on looks strange because of the absolute value. But whenever we become nervous, we can just take cases.
  • 109. Step 0: Finding Zeroes √ f(x) = x + |x| First, look at f by itself. We can tell that f(0) = 0 and that f(x) > 0 if x is posi ve.
  • 110. Step 0: Finding Zeroes √ f(x) = x + |x| First, look at f by itself. We can tell that f(0) = 0 and that f(x) > 0 if x is posi ve. Are there nega ve numbers which are zeroes for f?
  • 111. Step 0: Finding Zeroes √ f(x) = x + |x| First, look at f by itself. We can tell that f(0) = 0 and that f(x) > 0 if x is posi ve. Are there nega ve numbers which are zeroes for f? √ √ x + −x = 0 =⇒ −x = −x −x = x2 =⇒ x2 + x = 0 The only solu ons are x = 0 and x = −1.
  • 112. Step 0: Asymptotic behavior √ f(x) = x + |x| lim f(x) = ∞, because both terms tend to ∞. x→∞
  • 113. Step 0: Asymptotic behavior √ f(x) = x + |x| lim f(x) = ∞, because both terms tend to ∞. x→∞ lim f(x) is indeterminate of the form −∞ + ∞. It’s the same x→−∞ √ as lim (−y + y) y→+∞
  • 114. Step 0: Asymptotic behavior √ f(x) = x + |x| lim f(x) = ∞, because both terms tend to ∞. x→∞ lim f(x) is indeterminate of the form −∞ + ∞. It’s the same x→−∞ √ as lim (−y + y) y→+∞ √ √ √ y+y lim (−y + y) = lim ( y − y) · √ y→+∞ y→∞ y+y y − y2 = lim √ = −∞ y→∞ y+y
  • 115. Step 1: The derivative √ Remember, f(x) = x + |x|. To find f′ , first assume x > 0. Then d ( √ ) 1 f′ (x) = x+ x =1+ √ dx 2 x
  • 116. Step 1: The derivative √ Remember, f(x) = x + |x|. To find f′ , first assume x > 0. Then d ( √ ) 1 f′ (x) = x+ x =1+ √ dx 2 x No ce f′ (x) > 0 when x > 0 (so no cri cal points here)
  • 117. Step 1: The derivative √ Remember, f(x) = x + |x|. To find f′ , first assume x > 0. Then d ( √ ) 1 f′ (x) = x+ x =1+ √ dx 2 x No ce f′ (x) > 0 when x > 0 (so no cri cal points here) lim+ f′ (x) = ∞ (so 0 is a cri cal point) x→0
  • 118. Step 1: The derivative √ Remember, f(x) = x + |x|. To find f′ , first assume x > 0. Then d ( √ ) 1 f′ (x) = x+ x =1+ √ dx 2 x No ce f′ (x) > 0 when x > 0 (so no cri cal points here) lim+ f′ (x) = ∞ (so 0 is a cri cal point) x→0 lim f′ (x) = 1 (so the graph is asympto c to a line of slope 1) x→∞
  • 119. Step 1: The derivative √ Remember, f(x) = x + |x|. If x is nega ve, we have d ( √ ) 1 f′ (x) = x + −x = 1 − √ dx 2 −x No ce lim− f′ (x) = −∞ (other side of the cri cal point) x→0
  • 120. Step 1: The derivative √ Remember, f(x) = x + |x|. If x is nega ve, we have d ( √ ) 1 f′ (x) = x + −x = 1 − √ dx 2 −x No ce lim− f′ (x) = −∞ (other side of the cri cal point) x→0 lim f′ (x) = 1 (asympto c to a line of slope 1) x→−∞
  • 121. Step 1: The derivative √ Remember, f(x) = x + |x|. If x is nega ve, we have d ( √ ) 1 f′ (x) = x + −x = 1 − √ dx 2 −x No ce lim− f′ (x) = −∞ (other side of the cri cal point) x→0 lim f′ (x) = 1 (asympto c to a line of slope 1) x→−∞ ′ f (x) = 0 when 1 √ 1 1 1 1− √ = 0 =⇒ −x = =⇒ −x = =⇒ x = − 2 −x 2 4 4
  • 122. Step 1: Monotonicity  1 1 + √  if x > 0 ′ f (x) = 2 x 1 − √ 1  if x < 0 2 −x We can’t make a mul -factor sign chart because of the absolute value, but we can test points in between cri cal points. f′ (x) . f(x)
  • 123. Step 1: Monotonicity  1 1 + √  if x > 0 ′ f (x) = 2 x 1 − √ 1  if x < 0 2 −x We can’t make a mul -factor sign chart because of the absolute value, but we can test points in between cri cal points. 0 f′ (x) . −1 4 f(x)
  • 124. Step 1: Monotonicity  1 1 + √  if x > 0 ′ f (x) = 2 x 1 − √ 1  if x < 0 2 −x We can’t make a mul -factor sign chart because of the absolute value, but we can test points in between cri cal points. 0 ∞ f′ (x) . −1 4 0 f(x)
  • 125. Step 1: Monotonicity  1 1 + √  if x > 0 ′ f (x) = 2 x 1 − √ 1  if x < 0 2 −x We can’t make a mul -factor sign chart because of the absolute value, but we can test points in between cri cal points. + 0 ∞ f′ (x) . −1 4 0 f(x)
  • 126. Step 1: Monotonicity  1 1 + √  if x > 0 ′ f (x) = 2 x 1 − √ 1  if x < 0 2 −x We can’t make a mul -factor sign chart because of the absolute value, but we can test points in between cri cal points. + 0− ∞ f′ (x) . −4 0 1 f(x)
  • 127. Step 1: Monotonicity  1 1 + √  if x > 0 ′ f (x) = 2 x 1 − √ 1  if x < 0 2 −x We can’t make a mul -factor sign chart because of the absolute value, but we can test points in between cri cal points. + 0− ∞ + f′ (x) . −4 0 1 f(x)
  • 128. Step 1: Monotonicity  1 1 + √  if x > 0 ′ f (x) = 2 x 1 − √ 1  if x < 0 2 −x We can’t make a mul -factor sign chart because of the absolute value, but we can test points in between cri cal points. + 0− ∞ + f′ (x) . ↗ −4 0 1 f(x)
  • 129. Step 1: Monotonicity  1 1 + √  if x > 0 ′ f (x) = 2 x 1 − √ 1  if x < 0 2 −x We can’t make a mul -factor sign chart because of the absolute value, but we can test points in between cri cal points. + 0− ∞ + f′ (x) . ↗ −4 1↘ 0 f(x)
  • 130. Step 1: Monotonicity  1 1 + √  if x > 0 ′ f (x) = 2 x 1 − √ 1  if x < 0 2 −x We can’t make a mul -factor sign chart because of the absolute value, but we can test points in between cri cal points. + 0− ∞ + f′ (x) . ↗ −4 1↘ 0 ↗ f(x)
  • 131. Step 1: Monotonicity  1 1 + √  if x > 0 ′ f (x) = 2 x 1 − √ 1  if x < 0 2 −x We can’t make a mul -factor sign chart because of the absolute value, but we can test points in between cri cal points. + 0− ∞ + f′ (x) . ↗ −41↘ 0 ↗ f(x) max
  • 132. Step 1: Monotonicity  1 1 + √  if x > 0 ′ f (x) = 2 x 1 − √ 1  if x < 0 2 −x We can’t make a mul -factor sign chart because of the absolute value, but we can test points in between cri cal points. + 0− ∞ + f′ (x) . ↗ −41↘ 0 ↗ f(x) max min
  • 133. Step 2: Concavity ( ) d 1 1 If x > 0, then f′′ (x) = 1 + x−1/2 = − x−3/2 This is dx 2 4 nega ve whenever x > 0.
  • 134. Step 2: Concavity ( ) d 1 −1/2 1 If x > 0, then f′′ (x) = 1+ x = − x−3/2 This is dx 2 4 nega ve whenever x > 0. ( ) ′′ d 1 −1/2 1 If x < 0, then f (x) = 1 − (−x) = − (−x)−3/2 dx 2 4 which is also always nega ve for nega ve x.
  • 135. Step 2: Concavity ( ) d 1 −1/2 1 If x > 0, then f′′ (x) = 1+ x = − x−3/2 This is dx 2 4 nega ve whenever x > 0. ( ) ′′ d 1 −1/2 1 If x < 0, then f (x) = 1 − (−x) = − (−x)−3/2 dx 2 4 which is also always nega ve for nega ve x. 1 In other words, f′′ (x) = − |x|−3/2 . 4
  • 136. Step 2: Concavity ( ) d 1 −1/2 1 If x > 0, then f′′ (x) = 1+ x = − x−3/2 This is dx 2 4 nega ve whenever x > 0. ( ) ′′ d 1 −1/2 1 If x < 0, then f (x) = 1 − (−x) = − (−x)−3/2 dx 2 4 which is also always nega ve for nega ve x. 1 In other words, f′′ (x) = − |x|−3/2 . 4 Here is the sign chart: −− −∞ −− f′′ (x) . ⌢ 0 ⌢ f(x)
  • 137. Step 3: Synthesis Now we can put these things together. √ f(x) = x + |x| +1 + 0− ∞ + f′ +1 (x) . ↗ ↗ 1↘ 0 ↗ ↗monotonicity −∞ −− − −− 4 −∞ −− f′′ −∞ (x) ⌢ ⌢ 1 ⌢0 ⌢ ⌢concavity −∞ 0 4 0 +∞f(x) −1 −4 0 1 shape zero max min
  • 138. Step 3: Synthesis Now we can put these things together. √ f(x) = x + |x| +1 + 0− ∞ + f′ +1 (x) . ↗ ↗ 1↘ 0 ↗ ↗monotonicity −∞ −− − −− 4 −∞ −− f′′ −∞ (x) ⌢ ⌢ 1 ⌢0 ⌢ ⌢concavity −∞ 0 4 0 +∞f(x) −1 −4 0 1 shape zero max min
  • 139. Step 3: Synthesis Now we can put these things together. √ f(x) = x + |x| +1 + 0− ∞ + f′ +1 (x) . ↗ ↗ 1↘ 0 ↗ ↗monotonicity −∞ −− − −− 4 −∞ −− f′′ −∞ (x) ⌢ ⌢ 1 ⌢0 ⌢ ⌢concavity −∞ 0 4 0 +∞f(x) −1 −4 0 1 shape zero max min
  • 140. Step 3: Synthesis Now we can put these things together. √ f(x) = x + |x| +1 + 0− ∞ + f′ +1 (x) . ↗ ↗ 1↘ 0 ↗ ↗monotonicity −∞ −− − −− 4 −∞ −− f′′ −∞ (x) ⌢ ⌢ 1 ⌢0 ⌢ ⌢concavity −∞ 0 4 0 +∞f(x) −1 −4 0 1 shape zero max min
  • 141. Step 3: Synthesis Now we can put these things together. √ f(x) = x + |x| +1 + 0− ∞ + f′ +1 (x) . ↗ ↗ 1↘ 0 ↗ ↗monotonicity −∞ −− − −− 4 −∞ −− f′′ −∞ (x) ⌢ ⌢ 1 ⌢0 ⌢ ⌢concavity −∞ 0 4 0 +∞f(x) −1 −4 0 1 shape zero max min
  • 142. Graph √ f(x) = x + |x| . x 1 −∞0 4 0 +∞ x −1 −1 0 shape 4 zero max min
  • 143. Graph √ f(x) = x + |x| (−1, 0) . x 1 −∞0 4 0 +∞ x −1 −1 0 shape 4 zero max min
  • 144. Graph √ f(x) = x + |x| (− 1 , 1 ) 4 4 (−1, 0) . x 1 −∞0 4 0 +∞ x −1 −1 0 shape 4 zero max min
  • 145. Graph √ f(x) = x + |x| (− 1 , 1 ) 4 4 (−1, 0) . x (0, 0) 1 −∞0 4 0 +∞ x −1 −1 0 shape 4 zero max min
  • 146. Graph √ f(x) = x + |x| (− 1 , 1 ) 4 4 (−1, 0) . x (0, 0) 1 −∞0 4 0 +∞ x −1 −1 0 shape 4 zero max min
  • 147. Graph √ f(x) = x + |x| (− 1 , 1 ) 4 4 (−1, 0) . x (0, 0) 1 −∞0 4 0 +∞ x −1 −1 0 shape 4 zero max min
  • 148. Graph √ f(x) = x + |x| (− 1 , 1 ) 4 4 (−1, 0) . x (0, 0) 1 −∞0 4 0 +∞ x −1 −1 0 shape 4 zero max min
  • 149. Example with Horizontal Asymptotes Example Graph f(x) = xe−x 2
  • 150. Example with Horizontal Asymptotes Example Graph f(x) = xe−x 2 Before taking deriva ves, we no ce that f is odd, that f(0) = 0, and lim f(x) = 0 x→∞
  • 151. Step 1: −xMonotonicity 2 If f(x) = xe , then ( ) 2 f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x 2 2 ( √ )( √ ) −x2 = 1 − 2x 1 + 2x e . √0 √ 1− 2x 0 1/2 √ √ 1 + 2x − 0 1/2 0 f′ (x) √ √ − 1/2 1/2 f(x)
  • 152. Step 1: −xMonotonicity 2 If f(x) = xe , then ( ) 2 f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x 2 2 ( √ )( √ ) −x2 = 1 − 2x 1 + 2x e + . √0 √ 1− 2x 0 1/2 √ √ 1 + 2x − 0 1/2 0 f′ (x) √ √ − 1/2 1/2 f(x)
  • 153. Step 1: −xMonotonicity 2 If f(x) = xe , then ( ) 2 f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x 2 2 ( √ )( √ ) −x2 = 1 − 2x 1 + 2x e + + 0 . √ √ 1− 2x 0 1/2 √ √ 1 + 2x − 0 1/2 0 f′ (x) √ √ − 1/2 1/2 f(x)
  • 154. Step 1: −xMonotonicity 2 If f(x) = xe , then ( ) 2 f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x 2 2 ( √ )( √ ) −x2 = 1 − 2x 1 + 2x e + + 0 . √ − √ 1− 2x 0 1/2 √ √ 1 + 2x − 0 1/2 0 f′ (x) √ √ − 1/2 1/2 f(x)
  • 155. Step 1: −xMonotonicity 2 If f(x) = xe , then ( ) 2 f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x 2 2 ( √ )( √ ) −x2 = 1 − 2x 1 + 2x e + + 0 . √ − √ 1− 2x − 0 1/2 √ √ 1 + 2x − 0 1/2 0 f′ (x) √ √ − 1/2 1/2 f(x)
  • 156. Step 1: −xMonotonicity 2 If f(x) = xe , then ( ) 2 f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x 2 2 ( √ )( √ ) −x2 = 1 − 2x 1 + 2x e + + 0 . √ − √ 1− 2x − 0 + 1/2 √ √ 1 + 2x − 0 1/2 0 f′ (x) √ √ − 1/2 1/2 f(x)
  • 157. Step 1: −xMonotonicity 2 If f(x) = xe , then ( ) 2 f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x 2 2 ( √ )( √ ) −x2 = 1 − 2x 1 + 2x e + + 0 . √ − √ 1− 2x − 0 + 1/2 + √ √ 1 + 2x − 0 1/2 0 f′ (x) √ √ − 1/2 1/2 f(x)
  • 158. Step 1: −xMonotonicity 2 If f(x) = xe , then ( ) 2 f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x 2 2 ( √ )( √ ) −x2 = 1 − 2x 1 + 2x e + + 0 . √ − √ 1− 2x − 0 + 1/2 + √ √ 1 + 2x − − 0 1/2 0 f′ (x) √ √ − 1/2 1/2 f(x)
  • 159. Step 1: −xMonotonicity 2 If f(x) = xe , then ( ) 2 f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x 2 2 ( √ )( √ ) −x2 = 1 − 2x 1 + 2x e + + 0 . √ − √ 1− 2x − 0 + 1/2 + √ √ 1 + 2x − − 0 1/2 + 0 f′ (x) √ √ − 1/2 1/2 f(x)
  • 160. Step 1: −xMonotonicity 2 If f(x) = xe , then ( ) 2 f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x 2 2 ( √ )( √ ) −x2 = 1 − 2x 1 + 2x e + + 0 . √ − √ 1− 2x − 0 + 1/2 + √ √ 1 + 2x − − 0 1/2 + 0 − f′ (x) √ √ ↘ − 1/2 1/2 f(x)
  • 161. Step 1: −xMonotonicity 2 If f(x) = xe , then ( ) 2 f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x 2 2 ( √ )( √ ) −x2 = 1 − 2x 1 + 2x e + + 0 . √ − √ 1− 2x − 0 + 1/2 + √ √ 1 + 2x − − 0 1/2 + 0 − f′ (x) √ √ ↘ − 1/2 ↗ 1/2 f(x)
  • 162. Step 1: −xMonotonicity 2 If f(x) = xe , then ( ) 2 f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x 2 2 ( √ )( √ ) −x2 = 1 − 2x 1 + 2x e + + 0 . √ − √ 1− 2x − 0 + 1/2 + √ √ 1 + 2x − − 0 1/2 + 0 − f′ (x) √ √ ↘ − 1/2 ↗ 1/2 ↘ f(x)
  • 163. Step 1: −xMonotonicity 2 If f(x) = xe , then ( ) 2 f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x 2 2 ( √ )( √ ) −x2 = 1 − 2x 1 + 2x e + + 0 . √ − √ 1− 2x − 0 + 1/2 + √ √ 1 + 2x − − 0 1/2 + 0 − f′ (x) √ √ ↘ − 1/2 ↗ 1/2 ↘ f(x)
  • 164. Step 1: −xMonotonicity 2 If f(x) = xe , then ( ) 2 f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x 2 2 ( √ )( √ ) −x2 = 1 − 2x 1 + 2x e + + 0 . √ − √ 1− 2x − 0 + 1/2 + √ √ 1 + 2x − − 0 1/2 + 0 − f′ (x) √ √ ↘ − 1/2 ↗ 1/2 ↘ f(x) min
  • 165. Step 1: −xMonotonicity 2 If f(x) = xe , then ( ) 2 f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x 2 2 ( √ )( √ ) −x2 = 1 − 2x 1 + 2x e + + 0 . √ − √ 1− 2x − 0 + 1/2 + √ √ 1 + 2x − − 0 1/2 + 0 − f′ (x) √ √ ↘ − 1/2 ↗ 1/2 ↘ f(x) min max
  • 166. Step 2: Concavity ′ 2 −x 2 If f (x) = (1 − 2x )e , we know ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 = 2x(2x2 − 3)e−x 2 0. 2x 0 0 √ √ √ 2x − 3 0 3/2 √ √ √ 2x + 3 − 0 3/2 0 0 f′′ (x) √ √ − 3/2 0 3/2 f(x)
  • 167. Step 2: Concavity ′ 2 −x 2 If f (x) = (1 − 2x )e , we know ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 = 2x(2x2 − 3)e−x 2 − 0. 2x 0 0 √ √ √ 2x − 3 0 3/2 √ √ √ 2x + 3 − 0 3/2 0 0 f′′ (x) √ √ − 3/2 0 3/2 f(x)
  • 168. Step 2: Concavity ′ 2 −x 2 If f (x) = (1 − 2x )e , we know ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 = 2x(2x2 − 3)e−x 2 − − 0. 2x 0 0 √ √ √ 2x − 3 0 3/2 √ √ √ 2x + 3 − 0 3/2 0 0 f′′ (x) √ √ − 3/2 0 3/2 f(x)
  • 169. Step 2: Concavity ′ 2 −x 2 If f (x) = (1 − 2x )e , we know ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 = 2x(2x2 − 3)e−x 2 − − 0. + 2x 0 0 √ √ √ 2x − 3 0 3/2 √ √ √ 2x + 3 − 0 3/2 0 0 f′′ (x) √ √ − 3/2 0 3/2 f(x)
  • 170. Step 2: Concavity ′ 2 −x 2 If f (x) = (1 − 2x )e , we know ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 = 2x(2x2 − 3)e−x 2 − − 0. + + 2x 0 0 √ √ √ 2x − 3 0 3/2 √ √ √ 2x + 3 − 0 3/2 0 0 f′′ (x) √ √ − 3/2 0 3/2 f(x)
  • 171. Step 2: Concavity ′ 2 −x 2 If f (x) = (1 − 2x )e , we know ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 = 2x(2x2 − 3)e−x 2 − − 0. + + 2x 0 − 0 √ √ √ 2x − 3 0 3/2 √ √ √ 2x + 3 − 0 3/2 0 0 f′′ (x) √ √ − 3/2 0 3/2 f(x)
  • 172. Step 2: Concavity ′ 2 −x 2 If f (x) = (1 − 2x )e , we know ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 = 2x(2x2 − 3)e−x 2 − − 0. + + 2x 0 − − 0 √ √ √ 2x − 3 0 3/2 √ √ √ 2x + 3 − 0 3/2 0 0 f′′ (x) √ √ − 3/2 0 3/2 f(x)
  • 173. Step 2: Concavity ′ 2 −x 2 If f (x) = (1 − 2x )e , we know ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 = 2x(2x2 − 3)e−x 2 − − 0. + + 2x 0 − − − 0 √ √ √ 2x − 3 0 3/2 √ √ √ 2x + 3 − 0 3/2 0 0 f′′ (x) √ √ − 3/2 0 3/2 f(x)
  • 174. Step 2: Concavity ′ 2 −x 2 If f (x) = (1 − 2x )e , we know ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 = 2x(2x2 − 3)e−x 2 − − 0. + + 2x 0 − − − 0 + √ √ √ 2x − 3 0 3/2 √ √ √ 2x + 3 − 0 3/2 0 0 f′′ (x) √ √ − 3/2 0 3/2 f(x)
  • 175. Step 2: Concavity ′ 2 −x 2 If f (x) = (1 − 2x )e , we know ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 = 2x(2x2 − 3)e−x 2 − − 0. + + 2x 0 − − − 0 + √ √ √ 2x − 3 − 0 3/2 √ √ √ 2x + 3 − 0 3/2 0 0 f′′ (x) √ √ − 3/2 0 3/2 f(x)
  • 176. Step 2: Concavity ′ 2 −x 2 If f (x) = (1 − 2x )e , we know ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 = 2x(2x2 − 3)e−x 2 − − 0. + + 2x 0 − − − 0 + √ √ √ 2x − 3 − 0 + 3/2 √ √ √ 2x + 3 − 0 3/2 0 0 f′′ (x) √ √ − 3/2 0 3/2 f(x)
  • 177. Step 2: Concavity ′ 2 −x 2 If f (x) = (1 − 2x )e , we know ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 = 2x(2x2 − 3)e−x 2 − − 0. + + 2x 0 − − − 0 + √ √ √ 2x − 3 − 0 + + 3/2 √ √ √ 2x + 3 − 0 3/2 0 0 f′′ (x) √ √ − 3/2 0 3/2 f(x)
  • 178. Step 2: Concavity ′ 2 −x 2 If f (x) = (1 − 2x )e , we know ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 = 2x(2x2 − 3)e−x 2 − − 0. + + 2x 0 − − − 0 + √ √ √ 2x − 3 − 0 + + 3/2 + √ √ √ 2x + 3 − 0 3/2 0 0 f′′ (x) √ √ − 3/2 0 3/2 f(x)
  • 179. Step 2: Concavity ′ 2 −x 2 If f (x) = (1 − 2x )e , we know ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 = 2x(2x2 − 3)e−x 2 − − 0. + + 2x 0 − − − 0 + √ √ √ 2x − 3 − 0 + + 3/2 + √ √ √ 2x + 3 −− − 0 3/2 0 0 f′′ (x) √ √ − 3/2 0 3/2 f(x)
  • 180. Step 2: Concavity ′ 2 −x 2 If f (x) = (1 − 2x )e , we know ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 = 2x(2x2 − 3)e−x 2 − − 0. + + 2x 0 − − − 0 + √ √ √ 2x − 3 − 0 + + 3/2 + √ √ √ 2x + 3 −− − 0 3/2 ++ 0 0 f′′ (x) √ √ − 3/2 0 3/2 f(x)
  • 181. Step 2: Concavity ′ 2 −x 2 If f (x) = (1 − 2x )e , we know ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 = 2x(2x2 − 3)e−x 2 − − 0. + + 2x 0 − − − 0 + √ √ √ 2x − 3 − 0 + + 3/2 + √ √ √ 2x + 3 −− − 0 3/2 ++ 0 −− 0 f′′ (x) √ √ − 3/2 0 3/2 f(x)
  • 182. Step 2: Concavity ′ 2 −x 2 If f (x) = (1 − 2x )e , we know ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 = 2x(2x2 − 3)e−x 2 − − 0. + + 2x 0 − − − 0 + √ √ √ 2x − 3 − 0 + + 3/2 + √ √ √ 2x + 3 −− − 0 3/2 ++ 0 −− 0 ++ f′′ (x) ⌢ √ √ − 3/2 0 3/2 f(x)
  • 183. Step 2: Concavity ′ 2 −x 2 If f (x) = (1 − 2x )e , we know ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 = 2x(2x2 − 3)e−x 2 − − 0. + + 2x 0 − − − 0 + √ √ √ 2x − 3 − 0 + + 3/2 + √ √ √ 2x + 3 −− − 0 3/2 ++ 0 −− 0 ++ f′′ (x) ⌢ √ ⌣ √ − 3/2 0 3/2 f(x)
  • 184. Step 2: Concavity ′ 2 −x 2 If f (x) = (1 − 2x )e , we know ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 = 2x(2x2 − 3)e−x 2 − − 0. + + 2x 0 − − − 0 + √ √ √ 2x − 3 − 0 + + 3/2 + √ √ √ 2x + 3 −− − 0 3/2 ++ 0 −− 0 ++ f′′ (x) ⌢ √ ⌣ √ − 3/2 0 ⌢ 3/2 f(x)
  • 185. Step 2: Concavity ′ 2 −x 2 If f (x) = (1 − 2x )e , we know ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 = 2x(2x2 − 3)e−x 2 − − 0. + + 2x 0 − − − 0 + √ √ √ 2x − 3 − 0 + + 3/2 + √ √ √ 2x + 3 −− − 0 3/2 ++ 0 −− 0 ++ f′′ (x) ⌢ √ ⌣ √ − 3/2 0 ⌢ 3/2 ⌣ f(x)
  • 186. Step 2: Concavity ′ 2 −x 2 If f (x) = (1 − 2x )e , we know ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 = 2x(2x2 − 3)e−x 2 − − 0. + + 2x 0 − − − 0 + √ √ √ 2x − 3 − 0 + + 3/2 + √ √ √ 2x + 3 −− − 0 3/2 ++ 0 −− 0 ++ f′′ (x) ⌢ √ ⌣ √ − 3/2 0 ⌢ 3/2 ⌣ f(x)
  • 187. Step 2: Concavity ′ 2 −x 2 If f (x) = (1 − 2x )e , we know ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 = 2x(2x2 − 3)e−x 2 − − 0. + + 2x 0 − − − 0 + √ √ √ 2x − 3 − 0 + + 3/2 + √ √ √ 2x + 3 −− − 0 3/2 ++ 0 −− 0 ++ f′′ (x) ⌢ √ ⌣ √ − 3/2 0 ⌢ 3/2 ⌣ f(x) IP
  • 188. Step 2: Concavity ′ 2 −x 2 If f (x) = (1 − 2x )e , we know ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 = 2x(2x2 − 3)e−x 2 − − 0. + + 2x 0 − − − 0 + √ √ √ 2x − 3 − 0 + + 3/2 + √ √ √ 2x + 3 −− − 0 3/2 ++ 0 −− 0 ++ f′′ (x) ⌢ √ ⌣ √ − 3/2 0 ⌢ 3/2 ⌣ f(x) IP IP
  • 189. Step 2: Concavity ′ 2 −x 2 If f (x) = (1 − 2x )e , we know ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 = 2x(2x2 − 3)e−x 2 − − 0. + + 2x 0 − − − 0 + √ √ √ 2x − 3 − 0 + + 3/2 + √ √ √ 2x + 3 −− − 0 3/2 ++ 0 −− 0 ++ f′′ (x) ⌢ √ ⌣ √ − 3/2 0 ⌢ 3/2 ⌣ f(x) IP IP IP