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Sec on 4.1
    Maximum and Minimum Values
            V63.0121.011: Calculus I
          Professor Ma hew Leingang
                 New York University


                 April 4, 2011


.
Announcements

  Quiz 4 on Sec ons 3.3, 3.4, 3.5,
  and 3.7 next week (April 14/15)
  Quiz 5 on Sec ons 4.1–4.4
  April 28/29
  Final Exam Monday May 12,
  2:00–3:50pm
Objectives
  Understand and be able to
  explain the statement of the
  Extreme Value Theorem.
  Understand and be able to
  explain the statement of
  Fermat’s Theorem.
  Use the Closed Interval Method
  to find the extreme values of a
  func on defined on a closed
  interval.
Outline
 Introduc on

 The Extreme Value Theorem

 Fermat’s Theorem (not the last one)
    Tangent: Fermat’s Last Theorem

 The Closed Interval Method

 Examples
Optimize


.
Why go to the extremes?
  Ra onally speaking, it is
  advantageous to find the
  extreme values of a func on
  (maximize profit, minimize costs,
  etc.)




                                     Pierre-Louis Maupertuis
                                           (1698–1759)
Design




.
Why go to the extremes?
  Ra onally speaking, it is
  advantageous to find the
  extreme values of a func on
  (maximize profit, minimize costs,
  etc.)
  Many laws of science are
  derived from minimizing
  principles.

                                     Pierre-Louis Maupertuis
                                           (1698–1759)
Optics




.
Why go to the extremes?
  Ra onally speaking, it is
  advantageous to find the
  extreme values of a func on
  (maximize profit, minimize costs,
  etc.)
  Many laws of science are
  derived from minimizing
  principles.
  Maupertuis’ principle: “Ac on is
  minimized through the wisdom       Pierre-Louis Maupertuis
  of God.”                                 (1698–1759)
Outline
 Introduc on

 The Extreme Value Theorem

 Fermat’s Theorem (not the last one)
    Tangent: Fermat’s Last Theorem

 The Closed Interval Method

 Examples
Extreme points and values
        Defini on
        Let f have domain D.




                               .

Image credit: Patrick Q
Extreme points and values
        Defini on
        Let f have domain D.
              The func on f has an absolute maximum
              (or global maximum) (respec vely,
              absolute minimum) at c if f(c) ≥ f(x)
              (respec vely, f(c) ≤ f(x)) for all x in D



                                                          .

Image credit: Patrick Q
Extreme points and values
        Defini on
        Let f have domain D.
              The func on f has an absolute maximum
              (or global maximum) (respec vely,
              absolute minimum) at c if f(c) ≥ f(x)
              (respec vely, f(c) ≤ f(x)) for all x in D
              The number f(c) is called the maximum
              value (respec vely, minimum value) of f
              on D.
                                                          .

Image credit: Patrick Q
Extreme points and values
        Defini on
         Let f have domain D.
                      The func on f has an absolute maximum
                      (or global maximum) (respec vely,
                      absolute minimum) at c if f(c) ≥ f(x)
                      (respec vely, f(c) ≤ f(x)) for all x in D
                      The number f(c) is called the maximum
                      value (respec vely, minimum value) of f
                      on D.
                      An extremum is either a maximum or a        .
                      minimum. An extreme value is either a
                      maximum value or minimum value.
Image credit: Patrick Q
The Extreme Value Theorem
 Theorem (The Extreme Value
 Theorem)
 Let f be a func on which is
 con nuous on the closed
 interval [a, b]. Then f a ains
 an absolute maximum value
 f(c) and an absolute minimum
 value f(d) at numbers c and d
 in [a, b].
The Extreme Value Theorem
 Theorem (The Extreme Value
 Theorem)
 Let f be a func on which is
 con nuous on the closed
 interval [a, b]. Then f a ains
 an absolute maximum value
 f(c) and an absolute minimum
 value f(d) at numbers c and d     .
 in [a, b].                       a    b
The Extreme Value Theorem
 Theorem (The Extreme Value
 Theorem)                         maximum
                                      value
 Let f be a func on which is            f(c)
 con nuous on the closed
 interval [a, b]. Then f a ains
 an absolute maximum value
 f(c) and an absolute minimum
 value f(d) at numbers c and d                  .
                                               a        c
 in [a, b].                                         b
                                                        maximum
The Extreme Value Theorem
 Theorem (The Extreme Value
 Theorem)                         maximum
                                      value
 Let f be a func on which is            f(c)
 con nuous on the closed
 interval [a, b]. Then f a ains   minimum
 an absolute maximum value            value
 f(c) and an absolute minimum          f(d)
 value f(d) at numbers c and d                  .
                                               a        d    c
 in [a, b].                                                b
                                                    minimum maximum
No proof of EVT forthcoming


   This theorem is very hard to prove without using technical facts
   about con nuous func ons and closed intervals.
   But we can show the importance of each of the hypotheses.
Bad Example #1
 Example
 Consider the func on
         {
           x      0≤x<1
  f(x) =
           x − 2 1 ≤ x ≤ 2.
Bad Example #1
 Example
 Consider the func on
         {
           x      0≤x<1       .
  f(x) =                          |
           x − 2 1 ≤ x ≤ 2.       1
Bad Example #1
 Example
 Consider the func on
         {
           x      0≤x<1                            .
  f(x) =                                                |
           x − 2 1 ≤ x ≤ 2.                            1
 Then although values of f(x) get arbitrarily close to 1 and never
 bigger than 1, 1 is not the maximum value of f on [0, 1] because it is
 never achieved.
Bad Example #1
 Example
 Consider the func on
         {
           x      0≤x<1                            .
  f(x) =                                                |
           x − 2 1 ≤ x ≤ 2.                            1
 Then although values of f(x) get arbitrarily close to 1 and never
 bigger than 1, 1 is not the maximum value of f on [0, 1] because it is
 never achieved. This does not violate EVT because f is not
 con nuous.
Bad Example #2
 Example
 Consider the func on f(x) = x restricted to the interval [0, 1).




                                                  .         |
                                                           1
Bad Example #2
 Example
 Consider the func on f(x) = x restricted to the interval [0, 1).
      There is s ll no maximum
      value (values get
      arbitrarily close to 1 but
      do not achieve it).
                                                  .         |
                                                           1
Bad Example #2
 Example
 Consider the func on f(x) = x restricted to the interval [0, 1).
      There is s ll no maximum
      value (values get
      arbitrarily close to 1 but
      do not achieve it).
      This does not violate EVT
                                                  .         |
      because the domain is                                1
      not closed.
Final Bad Example
 Example
                      1
 The func on f(x) =     is con nuous on the closed interval [1, ∞).
                      x


           .
               1
Final Bad Example
 Example
                      1
 The func on f(x) =     is con nuous on the closed interval [1, ∞).
                      x


            .
                1

 There is no minimum value (values get arbitrarily close to 0 but do
 not achieve it).
Final Bad Example
 Example
                      1
 The func on f(x) =     is con nuous on the closed interval [1, ∞).
                      x


            .
                1

 There is no minimum value (values get arbitrarily close to 0 but do
 not achieve it). This does not violate EVT because the domain is not
 bounded.
Outline
 Introduc on

 The Extreme Value Theorem

 Fermat’s Theorem (not the last one)
    Tangent: Fermat’s Last Theorem

 The Closed Interval Method

 Examples
Local extrema
Defini on
   A func on f has a local
   maximum or rela ve maximum
   at c if f(c) ≥ f(x) when x is near
   c. This means that f(c) ≥ f(x)
   for all x in some open interval
   containing c.                        |.   |
                                        a    b
   Similarly, f has a local minimum
   at c if f(c) ≤ f(x) when x is near
   c.
Local extrema
Defini on
   A func on f has a local
   maximum or rela ve maximum
   at c if f(c) ≥ f(x) when x is near
   c. This means that f(c) ≥ f(x)
   for all x in some open interval
   containing c.                              |.    |
                                              a
                                            local   b
   Similarly, f has a local minimum     maximum
   at c if f(c) ≤ f(x) when x is near
   c.
Local extrema
Defini on
   A func on f has a local
   maximum or rela ve maximum
   at c if f(c) ≥ f(x) when x is near
   c. This means that f(c) ≥ f(x)
   for all x in some open interval
   containing c.                              |.        |
                                            local local b
                                              a
   Similarly, f has a local minimum     maximum minimum
   at c if f(c) ≤ f(x) when x is near
   c.
Local extrema
  A local extremum could be a
  global extremum, but not if
  there are more extreme values
  elsewhere.
  A global extremum could be a
  local extremum, but not if it is
  an endpoint.
                                           |.           |
                                           a           b
                                         local local and global
                                     maximum global max
                                                  min
Fermat’s Theorem
 Theorem (Fermat’s Theorem)


 Suppose f has a
 local extremum at c
 and f is
 differen able at c.
 Then f′ (c) = 0.                |.             |
                                a local   local b
                              maximum     minimum
Fermat’s Theorem
 Theorem (Fermat’s Theorem)


 Suppose f has a
 local extremum at c
 and f is
 differen able at c.
 Then f′ (c) = 0.                |.             |
                                a local   local b
                              maximum     minimum
Proof of Fermat’s Theorem
 Suppose that f has a local maximum at c.
Proof of Fermat’s Theorem
 Suppose that f has a local maximum at c.
     If x is slightly greater than c, f(x) ≤ f(c). This means
                 f(x) − f(c)
                             ≤0
                    x−c
Proof of Fermat’s Theorem
 Suppose that f has a local maximum at c.
     If x is slightly greater than c, f(x) ≤ f(c). This means
                 f(x) − f(c)             f(x) − f(c)
                             ≤ 0 =⇒ lim              ≤0
                    x−c             x→c+    x−c
Proof of Fermat’s Theorem
 Suppose that f has a local maximum at c.
     If x is slightly greater than c, f(x) ≤ f(c). This means
                 f(x) − f(c)             f(x) − f(c)
                             ≤ 0 =⇒ lim              ≤0
                    x−c             x→c+    x−c

      The same will be true on the other end: if x is slightly less than
      c, f(x) ≤ f(c). This means
                 f(x) − f(c)
                             ≥0
                    x−c
Proof of Fermat’s Theorem
 Suppose that f has a local maximum at c.
     If x is slightly greater than c, f(x) ≤ f(c). This means
                 f(x) − f(c)             f(x) − f(c)
                             ≤ 0 =⇒ lim              ≤0
                    x−c             x→c+    x−c

      The same will be true on the other end: if x is slightly less than
      c, f(x) ≤ f(c). This means
                 f(x) − f(c)             f(x) − f(c)
                             ≥ 0 =⇒ lim              ≥0
                    x−c             x→c−    x−c
Proof of Fermat’s Theorem
 Suppose that f has a local maximum at c.
     If x is slightly greater than c, f(x) ≤ f(c). This means
                  f(x) − f(c)             f(x) − f(c)
                              ≤ 0 =⇒ lim              ≤0
                     x−c             x→c+    x−c

      The same will be true on the other end: if x is slightly less than
      c, f(x) ≤ f(c). This means
                  f(x) − f(c)             f(x) − f(c)
                              ≥ 0 =⇒ lim              ≥0
                     x−c             x→c−    x−c
                                  f(x) − f(c)
      Since the limit f′ (c) = lim            exists, it must be 0.
                              x→c    x−c
Meet the Mathematician: Pierre de Fermat



     1601–1665
     Lawyer and number
     theorist
     Proved many theorems,
     didn’t quite prove his last
     one
Tangent: Fermat’s Last Theorem
 Plenty of solu ons to
 x2 + y2 = z2 among posi ve
 whole numbers (e.g., x = 3,
 y = 4, z = 5)
Tangent: Fermat’s Last Theorem
 Plenty of solu ons to
 x2 + y2 = z2 among posi ve
 whole numbers (e.g., x = 3,
 y = 4, z = 5)
 No solu ons to x3 + y3 = z3
 among posi ve whole numbers
Tangent: Fermat’s Last Theorem
 Plenty of solu ons to
 x2 + y2 = z2 among posi ve
 whole numbers (e.g., x = 3,
 y = 4, z = 5)
 No solu ons to x3 + y3 = z3
 among posi ve whole numbers
 Fermat claimed no solu ons to
 xn + yn = zn but didn’t write
 down his proof
Tangent: Fermat’s Last Theorem
 Plenty of solu ons to
 x2 + y2 = z2 among posi ve
 whole numbers (e.g., x = 3,
 y = 4, z = 5)
 No solu ons to x3 + y3 = z3
 among posi ve whole numbers
 Fermat claimed no solu ons to
 xn + yn = zn but didn’t write
 down his proof
 Not solved un l 1998!
 (Taylor–Wiles)
Outline
 Introduc on

 The Extreme Value Theorem

 Fermat’s Theorem (not the last one)
    Tangent: Fermat’s Last Theorem

 The Closed Interval Method

 Examples
Flowchart for placing extrema
Thanks to Fermat
Suppose f is a                      c is a
                       .
                     start
con nuous                         local max
func on on
the closed,
bounded              Is c an
                             no   Is f diff’ble   no   f is not
interval           endpoint?          at c?           diff at c
[a, b], and c is
a global
                     yes              yes
maximum
                   c = a or
point.                            f′ (c) = 0
                    c = b
The Closed Interval Method
 This means to find the maximum value of f on [a, b], we need to:
      Evaluate f at the endpoints a and b
      Evaluate f at the cri cal points or cri cal numbers x where
      either f′ (x) = 0 or f is not differen able at x.
      The points with the largest func on value are the global
      maximum points
      The points with the smallest or most nega ve func on value
      are the global minimum points.
Outline
 Introduc on

 The Extreme Value Theorem

 Fermat’s Theorem (not the last one)
    Tangent: Fermat’s Last Theorem

 The Closed Interval Method

 Examples
Extreme values of a linear function
 Example
 Find the extreme values of f(x) = 2x − 5 on [−1, 2].
Extreme values of a linear function
 Example
 Find the extreme values of f(x) = 2x − 5 on [−1, 2].

 Solu on
 Since f′ (x) = 2, which is never
 zero, we have no cri cal points
 and we need only inves gate
 the endpoints:
      f(−1) = 2(−1) − 5 = −7
      f(2) = 2(2) − 5 = −1
Extreme values of a linear function
 Example
 Find the extreme values of f(x) = 2x − 5 on [−1, 2].

 Solu on
                                     So
 Since f′ (x) = 2, which is never
 zero, we have no cri cal points          The absolute minimum
 and we need only inves gate              (point) is at −1; the
 the endpoints:                           minimum value is −7.
      f(−1) = 2(−1) − 5 = −7              The absolute maximum
                                          (point) is at 2; the
      f(2) = 2(2) − 5 = −1
                                          maximum value is −1.
Extreme values of a quadratic
function
 Example
 Find the extreme values of f(x) = x2 − 1 on [−1, 2].
Extreme values of a quadratic
function
 Example
 Find the extreme values of f(x) = x2 − 1 on [−1, 2].

 Solu on
 We have f′ (x) = 2x, which is zero when x = 0.
Extreme values of a quadratic
function
 Example
 Find the extreme values of f(x) = x2 − 1 on [−1, 2].

 Solu on
 We have f′ (x) = 2x, which is zero when x = 0. So our points to
 check are:
     f(−1) =
     f(0) =
     f(2) =
Extreme values of a quadratic
function
 Example
 Find the extreme values of f(x) = x2 − 1 on [−1, 2].

 Solu on
 We have f′ (x) = 2x, which is zero when x = 0. So our points to
 check are:
     f(−1) = 0
     f(0) =
     f(2) =
Extreme values of a quadratic
function
 Example
 Find the extreme values of f(x) = x2 − 1 on [−1, 2].

 Solu on
 We have f′ (x) = 2x, which is zero when x = 0. So our points to
 check are:
     f(−1) = 0
     f(0) = − 1
     f(2) =
Extreme values of a quadratic
function
 Example
 Find the extreme values of f(x) = x2 − 1 on [−1, 2].

 Solu on
 We have f′ (x) = 2x, which is zero when x = 0. So our points to
 check are:
     f(−1) = 0
     f(0) = − 1
     f(2) = 3
Extreme values of a quadratic
function
 Example
 Find the extreme values of f(x) = x2 − 1 on [−1, 2].

 Solu on
 We have f′ (x) = 2x, which is zero when x = 0. So our points to
 check are:
     f(−1) = 0
     f(0) = − 1 (absolute min)
     f(2) = 3
Extreme values of a quadratic
function
 Example
 Find the extreme values of f(x) = x2 − 1 on [−1, 2].

 Solu on
 We have f′ (x) = 2x, which is zero when x = 0. So our points to
 check are:
     f(−1) = 0
     f(0) = − 1 (absolute min)
     f(2) = 3 (absolute max)
Extreme values of a cubic function
 Example
 Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2].
Extreme values of a cubic function
 Example
 Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2].

 Solu on
 Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal points at x = 0
 and x = 1.
Extreme values of a cubic function
 Example
 Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2].

 Solu on
 Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal points at x = 0
 and x = 1. The values to check are
Extreme values of a cubic function
 Example
 Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2].

 Solu on
 Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal points at x = 0
 and x = 1. The values to check are
      f(−1) = − 4
Extreme values of a cubic function
 Example
 Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2].

 Solu on
 Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal points at x = 0
 and x = 1. The values to check are
      f(−1) = − 4
      f(0) = 1
Extreme values of a cubic function
 Example
 Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2].

 Solu on
 Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal points at x = 0
 and x = 1. The values to check are
      f(−1) = − 4
      f(0) = 1
      f(1) = 0
Extreme values of a cubic function
 Example
 Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2].

 Solu on
 Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal points at x = 0
 and x = 1. The values to check are
      f(−1) = − 4
      f(0) = 1
      f(1) = 0
      f(2) = 5
Extreme values of a cubic function
 Example
 Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2].

 Solu on
 Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal points at x = 0
 and x = 1. The values to check are
      f(−1) = − 4 (global min)
      f(0) = 1
      f(1) = 0
      f(2) = 5
Extreme values of a cubic function
 Example
 Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2].

 Solu on
 Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal points at x = 0
 and x = 1. The values to check are
      f(−1) = − 4 (global min)
      f(0) = 1
      f(1) = 0
      f(2) = 5 (global max)
Extreme values of a cubic function
 Example
 Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2].

 Solu on
 Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal points at x = 0
 and x = 1. The values to check are
      f(−1) = − 4 (global min)
      f(0) = 1 (local max)
      f(1) = 0
      f(2) = 5 (global max)
Extreme values of a cubic function
 Example
 Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2].

 Solu on
 Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal points at x = 0
 and x = 1. The values to check are
      f(−1) = − 4 (global min)
      f(0) = 1 (local max)
      f(1) = 0 (local min)
      f(2) = 5 (global max)
Extreme values of an algebraic function
 Example
 Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2].
Extreme values of an algebraic function
 Example
 Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2].

 Solu on
 Write f(x) = x5/3 + 2x2/3 .
Extreme values of an algebraic function
 Example
 Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2].

 Solu on
 Write f(x) = x5/3 + 2x2/3 . Then
                       5      4       1
               f′ (x) = x2/3 + x−1/3 = x−1/3 (5x + 4)
                       3      3       3
Extreme values of an algebraic function
 Example
 Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2].

 Solu on
 Write f(x) = x5/3 + 2x2/3 . Then
                       5      4       1
               f′ (x) = x2/3 + x−1/3 = x−1/3 (5x + 4)
                       3      3       3
 Thus f′ (−4/5) = 0 and f is not differen able at 0. Thus there are two
 cri cal points.
Extreme values of an algebraic function
 Example
 Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2].

 Solu on
 Write f(x) = x5/3 + 2x2/3 .
     f(−1) =
     f(−4/5) =
     f(0) =
     f(2) =
Extreme values of an algebraic function
 Example
 Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2].

 Solu on
 Write f(x) = x5/3 + 2x2/3 .
     f(−1) = 1
     f(−4/5) =
     f(0) =
     f(2) =
Extreme values of an algebraic function
 Example
 Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2].

 Solu on
 Write f(x) = x5/3 + 2x2/3 .
     f(−1) = 1
     f(−4/5) = 1.0341
     f(0) =
     f(2) =
Extreme values of an algebraic function
 Example
 Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2].

 Solu on
 Write f(x) = x5/3 + 2x2/3 .
     f(−1) = 1
     f(−4/5) = 1.0341
     f(0) = 0
     f(2) =
Extreme values of an algebraic function
 Example
 Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2].

 Solu on
 Write f(x) = x5/3 + 2x2/3 .
     f(−1) = 1
     f(−4/5) = 1.0341
     f(0) = 0
     f(2) = 6.3496
Extreme values of an algebraic function
 Example
 Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2].

 Solu on
 Write f(x) = x5/3 + 2x2/3 .
     f(−1) = 1
     f(−4/5) = 1.0341
     f(0) = 0 (absolute min)
     f(2) = 6.3496
Extreme values of an algebraic function
 Example
 Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2].

 Solu on
 Write f(x) = x5/3 + 2x2/3 .
     f(−1) = 1
     f(−4/5) = 1.0341
     f(0) = 0 (absolute min)
     f(2) = 6.3496 (absolute max)
Extreme values of an algebraic function
 Example
 Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2].

 Solu on
 Write f(x) = x5/3 + 2x2/3 .
     f(−1) = 1
     f(−4/5) = 1.0341 (rela ve max)
     f(0) = 0 (absolute min)
     f(2) = 6.3496 (absolute max)
Extreme values of another algebraic function
  Example
                                      √
  Find the extreme values of f(x) =       4 − x2 on [−2, 1].
Extreme values of another algebraic function
  Example
                                      √
  Find the extreme values of f(x) =       4 − x2 on [−2, 1].

  Solu on
                       x
  We have f′ (x) = − √      , which is zero when x = 0. (f is not
                     4 − x2
  differen able at ±2 as well.)
Extreme values of another algebraic function
  Example
                                      √
  Find the extreme values of f(x) =       4 − x2 on [−2, 1].

  Solu on
                       x
  We have f′ (x) = − √      , which is zero when x = 0. (f is not
                     4 − x2
  differen able at ±2 as well.) So our points to check are:
      f(−2) =
Extreme values of another algebraic function
  Example
                                      √
  Find the extreme values of f(x) =       4 − x2 on [−2, 1].

  Solu on
                       x
  We have f′ (x) = − √      , which is zero when x = 0. (f is not
                     4 − x2
  differen able at ±2 as well.) So our points to check are:
      f(−2) = 0
      f(0) =
Extreme values of another algebraic function
  Example
                                      √
  Find the extreme values of f(x) =       4 − x2 on [−2, 1].

  Solu on
                       x
  We have f′ (x) = − √      , which is zero when x = 0. (f is not
                     4 − x2
  differen able at ±2 as well.) So our points to check are:
      f(−2) = 0
      f(0) = 2
      f(1) =
Extreme values of another algebraic function
  Example
                                      √
  Find the extreme values of f(x) =       4 − x2 on [−2, 1].

  Solu on
                       x
  We have f′ (x) = − √      , which is zero when x = 0. (f is not
                     4 − x2
  differen able at ±2 as well.) So our points to check are:
      f(−2) = 0
      f(0) = 2
             √
      f(1) = 3
Extreme values of another algebraic function
  Example
                                      √
  Find the extreme values of f(x) =       4 − x2 on [−2, 1].

  Solu on
                       x
  We have f′ (x) = − √      , which is zero when x = 0. (f is not
                     4 − x2
  differen able at ±2 as well.) So our points to check are:
      f(−2) = 0 (absolute min)
      f(0) = 2
             √
      f(1) = 3
Extreme values of another algebraic function
  Example
                                      √
  Find the extreme values of f(x) =       4 − x2 on [−2, 1].

  Solu on
                        x
  We have f′ (x) = − √       , which is zero when x = 0. (f is not
                      4 − x2
  differen able at ±2 as well.) So our points to check are:
      f(−2) = 0 (absolute min)
      f(0) = 2 (absolute max)
             √
      f(1) = 3
Summary

  The Extreme Value Theorem: a con nuous func on on a closed
  interval must achieve its max and min
  Fermat’s Theorem: local extrema are cri cal points
  The Closed Interval Method: an algorithm for finding global
  extrema
  Show your work unless you want to end up like Fermat!

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Lesson 18: Maximum and Minimum Values (slides)

  • 1. Sec on 4.1 Maximum and Minimum Values V63.0121.011: Calculus I Professor Ma hew Leingang New York University April 4, 2011 .
  • 2. Announcements Quiz 4 on Sec ons 3.3, 3.4, 3.5, and 3.7 next week (April 14/15) Quiz 5 on Sec ons 4.1–4.4 April 28/29 Final Exam Monday May 12, 2:00–3:50pm
  • 3. Objectives Understand and be able to explain the statement of the Extreme Value Theorem. Understand and be able to explain the statement of Fermat’s Theorem. Use the Closed Interval Method to find the extreme values of a func on defined on a closed interval.
  • 4. Outline Introduc on The Extreme Value Theorem Fermat’s Theorem (not the last one) Tangent: Fermat’s Last Theorem The Closed Interval Method Examples
  • 6. Why go to the extremes? Ra onally speaking, it is advantageous to find the extreme values of a func on (maximize profit, minimize costs, etc.) Pierre-Louis Maupertuis (1698–1759)
  • 8. Why go to the extremes? Ra onally speaking, it is advantageous to find the extreme values of a func on (maximize profit, minimize costs, etc.) Many laws of science are derived from minimizing principles. Pierre-Louis Maupertuis (1698–1759)
  • 10. Why go to the extremes? Ra onally speaking, it is advantageous to find the extreme values of a func on (maximize profit, minimize costs, etc.) Many laws of science are derived from minimizing principles. Maupertuis’ principle: “Ac on is minimized through the wisdom Pierre-Louis Maupertuis of God.” (1698–1759)
  • 11. Outline Introduc on The Extreme Value Theorem Fermat’s Theorem (not the last one) Tangent: Fermat’s Last Theorem The Closed Interval Method Examples
  • 12. Extreme points and values Defini on Let f have domain D. . Image credit: Patrick Q
  • 13. Extreme points and values Defini on Let f have domain D. The func on f has an absolute maximum (or global maximum) (respec vely, absolute minimum) at c if f(c) ≥ f(x) (respec vely, f(c) ≤ f(x)) for all x in D . Image credit: Patrick Q
  • 14. Extreme points and values Defini on Let f have domain D. The func on f has an absolute maximum (or global maximum) (respec vely, absolute minimum) at c if f(c) ≥ f(x) (respec vely, f(c) ≤ f(x)) for all x in D The number f(c) is called the maximum value (respec vely, minimum value) of f on D. . Image credit: Patrick Q
  • 15. Extreme points and values Defini on Let f have domain D. The func on f has an absolute maximum (or global maximum) (respec vely, absolute minimum) at c if f(c) ≥ f(x) (respec vely, f(c) ≤ f(x)) for all x in D The number f(c) is called the maximum value (respec vely, minimum value) of f on D. An extremum is either a maximum or a . minimum. An extreme value is either a maximum value or minimum value. Image credit: Patrick Q
  • 16. The Extreme Value Theorem Theorem (The Extreme Value Theorem) Let f be a func on which is con nuous on the closed interval [a, b]. Then f a ains an absolute maximum value f(c) and an absolute minimum value f(d) at numbers c and d in [a, b].
  • 17. The Extreme Value Theorem Theorem (The Extreme Value Theorem) Let f be a func on which is con nuous on the closed interval [a, b]. Then f a ains an absolute maximum value f(c) and an absolute minimum value f(d) at numbers c and d . in [a, b]. a b
  • 18. The Extreme Value Theorem Theorem (The Extreme Value Theorem) maximum value Let f be a func on which is f(c) con nuous on the closed interval [a, b]. Then f a ains an absolute maximum value f(c) and an absolute minimum value f(d) at numbers c and d . a c in [a, b]. b maximum
  • 19. The Extreme Value Theorem Theorem (The Extreme Value Theorem) maximum value Let f be a func on which is f(c) con nuous on the closed interval [a, b]. Then f a ains minimum an absolute maximum value value f(c) and an absolute minimum f(d) value f(d) at numbers c and d . a d c in [a, b]. b minimum maximum
  • 20. No proof of EVT forthcoming This theorem is very hard to prove without using technical facts about con nuous func ons and closed intervals. But we can show the importance of each of the hypotheses.
  • 21. Bad Example #1 Example Consider the func on { x 0≤x<1 f(x) = x − 2 1 ≤ x ≤ 2.
  • 22. Bad Example #1 Example Consider the func on { x 0≤x<1 . f(x) = | x − 2 1 ≤ x ≤ 2. 1
  • 23. Bad Example #1 Example Consider the func on { x 0≤x<1 . f(x) = | x − 2 1 ≤ x ≤ 2. 1 Then although values of f(x) get arbitrarily close to 1 and never bigger than 1, 1 is not the maximum value of f on [0, 1] because it is never achieved.
  • 24. Bad Example #1 Example Consider the func on { x 0≤x<1 . f(x) = | x − 2 1 ≤ x ≤ 2. 1 Then although values of f(x) get arbitrarily close to 1 and never bigger than 1, 1 is not the maximum value of f on [0, 1] because it is never achieved. This does not violate EVT because f is not con nuous.
  • 25. Bad Example #2 Example Consider the func on f(x) = x restricted to the interval [0, 1). . | 1
  • 26. Bad Example #2 Example Consider the func on f(x) = x restricted to the interval [0, 1). There is s ll no maximum value (values get arbitrarily close to 1 but do not achieve it). . | 1
  • 27. Bad Example #2 Example Consider the func on f(x) = x restricted to the interval [0, 1). There is s ll no maximum value (values get arbitrarily close to 1 but do not achieve it). This does not violate EVT . | because the domain is 1 not closed.
  • 28. Final Bad Example Example 1 The func on f(x) = is con nuous on the closed interval [1, ∞). x . 1
  • 29. Final Bad Example Example 1 The func on f(x) = is con nuous on the closed interval [1, ∞). x . 1 There is no minimum value (values get arbitrarily close to 0 but do not achieve it).
  • 30. Final Bad Example Example 1 The func on f(x) = is con nuous on the closed interval [1, ∞). x . 1 There is no minimum value (values get arbitrarily close to 0 but do not achieve it). This does not violate EVT because the domain is not bounded.
  • 31. Outline Introduc on The Extreme Value Theorem Fermat’s Theorem (not the last one) Tangent: Fermat’s Last Theorem The Closed Interval Method Examples
  • 32. Local extrema Defini on A func on f has a local maximum or rela ve maximum at c if f(c) ≥ f(x) when x is near c. This means that f(c) ≥ f(x) for all x in some open interval containing c. |. | a b Similarly, f has a local minimum at c if f(c) ≤ f(x) when x is near c.
  • 33. Local extrema Defini on A func on f has a local maximum or rela ve maximum at c if f(c) ≥ f(x) when x is near c. This means that f(c) ≥ f(x) for all x in some open interval containing c. |. | a local b Similarly, f has a local minimum maximum at c if f(c) ≤ f(x) when x is near c.
  • 34. Local extrema Defini on A func on f has a local maximum or rela ve maximum at c if f(c) ≥ f(x) when x is near c. This means that f(c) ≥ f(x) for all x in some open interval containing c. |. | local local b a Similarly, f has a local minimum maximum minimum at c if f(c) ≤ f(x) when x is near c.
  • 35. Local extrema A local extremum could be a global extremum, but not if there are more extreme values elsewhere. A global extremum could be a local extremum, but not if it is an endpoint. |. | a b local local and global maximum global max min
  • 36. Fermat’s Theorem Theorem (Fermat’s Theorem) Suppose f has a local extremum at c and f is differen able at c. Then f′ (c) = 0. |. | a local local b maximum minimum
  • 37. Fermat’s Theorem Theorem (Fermat’s Theorem) Suppose f has a local extremum at c and f is differen able at c. Then f′ (c) = 0. |. | a local local b maximum minimum
  • 38. Proof of Fermat’s Theorem Suppose that f has a local maximum at c.
  • 39. Proof of Fermat’s Theorem Suppose that f has a local maximum at c. If x is slightly greater than c, f(x) ≤ f(c). This means f(x) − f(c) ≤0 x−c
  • 40. Proof of Fermat’s Theorem Suppose that f has a local maximum at c. If x is slightly greater than c, f(x) ≤ f(c). This means f(x) − f(c) f(x) − f(c) ≤ 0 =⇒ lim ≤0 x−c x→c+ x−c
  • 41. Proof of Fermat’s Theorem Suppose that f has a local maximum at c. If x is slightly greater than c, f(x) ≤ f(c). This means f(x) − f(c) f(x) − f(c) ≤ 0 =⇒ lim ≤0 x−c x→c+ x−c The same will be true on the other end: if x is slightly less than c, f(x) ≤ f(c). This means f(x) − f(c) ≥0 x−c
  • 42. Proof of Fermat’s Theorem Suppose that f has a local maximum at c. If x is slightly greater than c, f(x) ≤ f(c). This means f(x) − f(c) f(x) − f(c) ≤ 0 =⇒ lim ≤0 x−c x→c+ x−c The same will be true on the other end: if x is slightly less than c, f(x) ≤ f(c). This means f(x) − f(c) f(x) − f(c) ≥ 0 =⇒ lim ≥0 x−c x→c− x−c
  • 43. Proof of Fermat’s Theorem Suppose that f has a local maximum at c. If x is slightly greater than c, f(x) ≤ f(c). This means f(x) − f(c) f(x) − f(c) ≤ 0 =⇒ lim ≤0 x−c x→c+ x−c The same will be true on the other end: if x is slightly less than c, f(x) ≤ f(c). This means f(x) − f(c) f(x) − f(c) ≥ 0 =⇒ lim ≥0 x−c x→c− x−c f(x) − f(c) Since the limit f′ (c) = lim exists, it must be 0. x→c x−c
  • 44. Meet the Mathematician: Pierre de Fermat 1601–1665 Lawyer and number theorist Proved many theorems, didn’t quite prove his last one
  • 45. Tangent: Fermat’s Last Theorem Plenty of solu ons to x2 + y2 = z2 among posi ve whole numbers (e.g., x = 3, y = 4, z = 5)
  • 46. Tangent: Fermat’s Last Theorem Plenty of solu ons to x2 + y2 = z2 among posi ve whole numbers (e.g., x = 3, y = 4, z = 5) No solu ons to x3 + y3 = z3 among posi ve whole numbers
  • 47. Tangent: Fermat’s Last Theorem Plenty of solu ons to x2 + y2 = z2 among posi ve whole numbers (e.g., x = 3, y = 4, z = 5) No solu ons to x3 + y3 = z3 among posi ve whole numbers Fermat claimed no solu ons to xn + yn = zn but didn’t write down his proof
  • 48. Tangent: Fermat’s Last Theorem Plenty of solu ons to x2 + y2 = z2 among posi ve whole numbers (e.g., x = 3, y = 4, z = 5) No solu ons to x3 + y3 = z3 among posi ve whole numbers Fermat claimed no solu ons to xn + yn = zn but didn’t write down his proof Not solved un l 1998! (Taylor–Wiles)
  • 49. Outline Introduc on The Extreme Value Theorem Fermat’s Theorem (not the last one) Tangent: Fermat’s Last Theorem The Closed Interval Method Examples
  • 50. Flowchart for placing extrema Thanks to Fermat Suppose f is a c is a . start con nuous local max func on on the closed, bounded Is c an no Is f diff’ble no f is not interval endpoint? at c? diff at c [a, b], and c is a global yes yes maximum c = a or point. f′ (c) = 0 c = b
  • 51. The Closed Interval Method This means to find the maximum value of f on [a, b], we need to: Evaluate f at the endpoints a and b Evaluate f at the cri cal points or cri cal numbers x where either f′ (x) = 0 or f is not differen able at x. The points with the largest func on value are the global maximum points The points with the smallest or most nega ve func on value are the global minimum points.
  • 52. Outline Introduc on The Extreme Value Theorem Fermat’s Theorem (not the last one) Tangent: Fermat’s Last Theorem The Closed Interval Method Examples
  • 53. Extreme values of a linear function Example Find the extreme values of f(x) = 2x − 5 on [−1, 2].
  • 54. Extreme values of a linear function Example Find the extreme values of f(x) = 2x − 5 on [−1, 2]. Solu on Since f′ (x) = 2, which is never zero, we have no cri cal points and we need only inves gate the endpoints: f(−1) = 2(−1) − 5 = −7 f(2) = 2(2) − 5 = −1
  • 55. Extreme values of a linear function Example Find the extreme values of f(x) = 2x − 5 on [−1, 2]. Solu on So Since f′ (x) = 2, which is never zero, we have no cri cal points The absolute minimum and we need only inves gate (point) is at −1; the the endpoints: minimum value is −7. f(−1) = 2(−1) − 5 = −7 The absolute maximum (point) is at 2; the f(2) = 2(2) − 5 = −1 maximum value is −1.
  • 56. Extreme values of a quadratic function Example Find the extreme values of f(x) = x2 − 1 on [−1, 2].
  • 57. Extreme values of a quadratic function Example Find the extreme values of f(x) = x2 − 1 on [−1, 2]. Solu on We have f′ (x) = 2x, which is zero when x = 0.
  • 58. Extreme values of a quadratic function Example Find the extreme values of f(x) = x2 − 1 on [−1, 2]. Solu on We have f′ (x) = 2x, which is zero when x = 0. So our points to check are: f(−1) = f(0) = f(2) =
  • 59. Extreme values of a quadratic function Example Find the extreme values of f(x) = x2 − 1 on [−1, 2]. Solu on We have f′ (x) = 2x, which is zero when x = 0. So our points to check are: f(−1) = 0 f(0) = f(2) =
  • 60. Extreme values of a quadratic function Example Find the extreme values of f(x) = x2 − 1 on [−1, 2]. Solu on We have f′ (x) = 2x, which is zero when x = 0. So our points to check are: f(−1) = 0 f(0) = − 1 f(2) =
  • 61. Extreme values of a quadratic function Example Find the extreme values of f(x) = x2 − 1 on [−1, 2]. Solu on We have f′ (x) = 2x, which is zero when x = 0. So our points to check are: f(−1) = 0 f(0) = − 1 f(2) = 3
  • 62. Extreme values of a quadratic function Example Find the extreme values of f(x) = x2 − 1 on [−1, 2]. Solu on We have f′ (x) = 2x, which is zero when x = 0. So our points to check are: f(−1) = 0 f(0) = − 1 (absolute min) f(2) = 3
  • 63. Extreme values of a quadratic function Example Find the extreme values of f(x) = x2 − 1 on [−1, 2]. Solu on We have f′ (x) = 2x, which is zero when x = 0. So our points to check are: f(−1) = 0 f(0) = − 1 (absolute min) f(2) = 3 (absolute max)
  • 64. Extreme values of a cubic function Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2].
  • 65. Extreme values of a cubic function Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solu on Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal points at x = 0 and x = 1.
  • 66. Extreme values of a cubic function Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solu on Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal points at x = 0 and x = 1. The values to check are
  • 67. Extreme values of a cubic function Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solu on Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal points at x = 0 and x = 1. The values to check are f(−1) = − 4
  • 68. Extreme values of a cubic function Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solu on Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal points at x = 0 and x = 1. The values to check are f(−1) = − 4 f(0) = 1
  • 69. Extreme values of a cubic function Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solu on Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal points at x = 0 and x = 1. The values to check are f(−1) = − 4 f(0) = 1 f(1) = 0
  • 70. Extreme values of a cubic function Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solu on Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal points at x = 0 and x = 1. The values to check are f(−1) = − 4 f(0) = 1 f(1) = 0 f(2) = 5
  • 71. Extreme values of a cubic function Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solu on Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal points at x = 0 and x = 1. The values to check are f(−1) = − 4 (global min) f(0) = 1 f(1) = 0 f(2) = 5
  • 72. Extreme values of a cubic function Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solu on Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal points at x = 0 and x = 1. The values to check are f(−1) = − 4 (global min) f(0) = 1 f(1) = 0 f(2) = 5 (global max)
  • 73. Extreme values of a cubic function Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solu on Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal points at x = 0 and x = 1. The values to check are f(−1) = − 4 (global min) f(0) = 1 (local max) f(1) = 0 f(2) = 5 (global max)
  • 74. Extreme values of a cubic function Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solu on Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal points at x = 0 and x = 1. The values to check are f(−1) = − 4 (global min) f(0) = 1 (local max) f(1) = 0 (local min) f(2) = 5 (global max)
  • 75. Extreme values of an algebraic function Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2].
  • 76. Extreme values of an algebraic function Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solu on Write f(x) = x5/3 + 2x2/3 .
  • 77. Extreme values of an algebraic function Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solu on Write f(x) = x5/3 + 2x2/3 . Then 5 4 1 f′ (x) = x2/3 + x−1/3 = x−1/3 (5x + 4) 3 3 3
  • 78. Extreme values of an algebraic function Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solu on Write f(x) = x5/3 + 2x2/3 . Then 5 4 1 f′ (x) = x2/3 + x−1/3 = x−1/3 (5x + 4) 3 3 3 Thus f′ (−4/5) = 0 and f is not differen able at 0. Thus there are two cri cal points.
  • 79. Extreme values of an algebraic function Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solu on Write f(x) = x5/3 + 2x2/3 . f(−1) = f(−4/5) = f(0) = f(2) =
  • 80. Extreme values of an algebraic function Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solu on Write f(x) = x5/3 + 2x2/3 . f(−1) = 1 f(−4/5) = f(0) = f(2) =
  • 81. Extreme values of an algebraic function Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solu on Write f(x) = x5/3 + 2x2/3 . f(−1) = 1 f(−4/5) = 1.0341 f(0) = f(2) =
  • 82. Extreme values of an algebraic function Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solu on Write f(x) = x5/3 + 2x2/3 . f(−1) = 1 f(−4/5) = 1.0341 f(0) = 0 f(2) =
  • 83. Extreme values of an algebraic function Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solu on Write f(x) = x5/3 + 2x2/3 . f(−1) = 1 f(−4/5) = 1.0341 f(0) = 0 f(2) = 6.3496
  • 84. Extreme values of an algebraic function Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solu on Write f(x) = x5/3 + 2x2/3 . f(−1) = 1 f(−4/5) = 1.0341 f(0) = 0 (absolute min) f(2) = 6.3496
  • 85. Extreme values of an algebraic function Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solu on Write f(x) = x5/3 + 2x2/3 . f(−1) = 1 f(−4/5) = 1.0341 f(0) = 0 (absolute min) f(2) = 6.3496 (absolute max)
  • 86. Extreme values of an algebraic function Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solu on Write f(x) = x5/3 + 2x2/3 . f(−1) = 1 f(−4/5) = 1.0341 (rela ve max) f(0) = 0 (absolute min) f(2) = 6.3496 (absolute max)
  • 87. Extreme values of another algebraic function Example √ Find the extreme values of f(x) = 4 − x2 on [−2, 1].
  • 88. Extreme values of another algebraic function Example √ Find the extreme values of f(x) = 4 − x2 on [−2, 1]. Solu on x We have f′ (x) = − √ , which is zero when x = 0. (f is not 4 − x2 differen able at ±2 as well.)
  • 89. Extreme values of another algebraic function Example √ Find the extreme values of f(x) = 4 − x2 on [−2, 1]. Solu on x We have f′ (x) = − √ , which is zero when x = 0. (f is not 4 − x2 differen able at ±2 as well.) So our points to check are: f(−2) =
  • 90. Extreme values of another algebraic function Example √ Find the extreme values of f(x) = 4 − x2 on [−2, 1]. Solu on x We have f′ (x) = − √ , which is zero when x = 0. (f is not 4 − x2 differen able at ±2 as well.) So our points to check are: f(−2) = 0 f(0) =
  • 91. Extreme values of another algebraic function Example √ Find the extreme values of f(x) = 4 − x2 on [−2, 1]. Solu on x We have f′ (x) = − √ , which is zero when x = 0. (f is not 4 − x2 differen able at ±2 as well.) So our points to check are: f(−2) = 0 f(0) = 2 f(1) =
  • 92. Extreme values of another algebraic function Example √ Find the extreme values of f(x) = 4 − x2 on [−2, 1]. Solu on x We have f′ (x) = − √ , which is zero when x = 0. (f is not 4 − x2 differen able at ±2 as well.) So our points to check are: f(−2) = 0 f(0) = 2 √ f(1) = 3
  • 93. Extreme values of another algebraic function Example √ Find the extreme values of f(x) = 4 − x2 on [−2, 1]. Solu on x We have f′ (x) = − √ , which is zero when x = 0. (f is not 4 − x2 differen able at ±2 as well.) So our points to check are: f(−2) = 0 (absolute min) f(0) = 2 √ f(1) = 3
  • 94. Extreme values of another algebraic function Example √ Find the extreme values of f(x) = 4 − x2 on [−2, 1]. Solu on x We have f′ (x) = − √ , which is zero when x = 0. (f is not 4 − x2 differen able at ±2 as well.) So our points to check are: f(−2) = 0 (absolute min) f(0) = 2 (absolute max) √ f(1) = 3
  • 95. Summary The Extreme Value Theorem: a con nuous func on on a closed interval must achieve its max and min Fermat’s Theorem: local extrema are cri cal points The Closed Interval Method: an algorithm for finding global extrema Show your work unless you want to end up like Fermat!