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Lesson 19: Curve Sketching
 

Lesson 19: Curve Sketching

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    Lesson 19: Curve Sketching Lesson 19: Curve Sketching Presentation Transcript

    • Section 4.4 Curve Sketching V63.0121.006/016, Calculus I New York University April 1, 2010 . . . . . .
    • Second-chance Midterm: Tomorrow in Recitation 12 free-response questions, no multiple choice Covers all sections so far, through today Your score on this exam will replace your midterm score . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 2 / 47
    • . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 3 / 47
    • Quiz 3 tomorrow in recitation Section 2.6: implicit differentiation Section 2.8: linear approximation and differentials Section 3.1: exponential functions Section 3.2: logarithms Section 3.3: derivatives of logarithmic and exponential functions Section 3.4: exponential growth and decay Section 3.5: inverse trigonometric functions . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 4 / 47
    • Outline The Procedure Simple examples A cubic function A quartic function More Examples Points of nondifferentiability Horizontal asymptotes Vertical asymptotes Trigonometric and polynomial together Logarithmic . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 5 / 47
    • Objective Given a function, graph it completely, indicating zeroes asymptotes if applicable critical points local/global max/min inflection points . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 6 / 47
    • Objective Given a function, graph it completely, indicating zeroes asymptotes if applicable critical points local/global max/min inflection points . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 6 / 47
    • 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 Here f(x) = x3 + x2 , and f′ (x) = 3x2 + 2x. f .(x) .′ (x) f . . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 7 / 47
    • 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 Here f(x) = x3 + x2 , f′ (x) = 3x2 + 2x, and f′′ (x) = 6x + 2. .′′ (x) f f .(x) .′ (x) f . . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 8 / 47
    • Graphing Checklist To graph a function f, follow this plan: 0. Find when f is positive, negative, zero, not defined. 1. Find f′ and form its sign chart. Conclude information about increasing/decreasing and local max/min. 2. Find f′′ and form its sign chart. Conclude concave up/concave down and inflection. 3. Put together a big chart to assemble monotonicity and concavity data 4. Graph! . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 9 / 47
    • Outline The Procedure Simple examples A cubic function A quartic function More Examples Points of nondifferentiability Horizontal asymptotes Vertical asymptotes Trigonometric and polynomial together Logarithmic . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 10 / 47
    • Graphing a cubic Example Graph f(x) = 2x3 − 3x2 − 12x. . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 11 / 47
    • 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 quadratic, 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. . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 11 / 47
    • 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: . . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 12 / 47
    • 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: . . . −2 x 2 . . x . +1 − . 1 .′ (x) f . . − . 1 2 . f .(x) . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 12 / 47
    • 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: − . − . . . . + . −2 x 2 . . x . +1 − . 1 .′ (x) f . . − . 1 2 . f .(x) . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 12 / 47
    • 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: − . − . . . . + . −2 x 2 . − . . . + . + x . +1 − . 1 .′ (x) f . . − . 1 2 . f .(x) . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 12 / 47
    • 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: − . − . . . . + . −2 x 2 . − . . . + . + x . +1 − . 1 . . + .′ (x) f . − . 1 2 . f .(x) . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 12 / 47
    • 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: − . − . . . . + . −2 x 2 . − . . . + . + x . +1 − . 1 . . + − . .′ (x) f . − . 1 2 . f .(x) . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 12 / 47
    • 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: − . − . . . . + . −2 x 2 . − . . . + . + x . +1 − . 1 . . + − . . + .′ (x) f . − . 1 2 . f .(x) . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 12 / 47
    • 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: − . − . . . . + . −2 x 2 . − . . . + . + x . +1 − . 1 . . + − . . + .′ (x) f . ↗− . . 1 2 . f .(x) . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 12 / 47
    • 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: − . − . . . . + . −2 x 2 . − . . . + . + x . +1 − . 1 . . + − . . + .′ (x) f . ↗− . . 1 ↘ . 2 . f .(x) . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 12 / 47
    • 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: − . − . . . . + . −2 x 2 . − . . . + . + x . +1 − . 1 . . + − . . + .′ (x) f . ↗− . . 1 ↘ . 2 . ↗ . f .(x) . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 12 / 47
    • 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: − . − . . . . + . −2 x 2 . − . . . + . + x . +1 − . 1 . . + − . . + .′ (x) f . ↗− . . 1 ↘ . 2 . ↗ . f .(x) m . ax . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 12 / 47
    • 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: − . − . . . . + . −2 x 2 . − . . . + . + x . +1 − . 1 . . + − . . + .′ (x) f . ↗− . . 1 ↘ . 2 . ↗ . f .(x) m . ax m . in . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 12 / 47
    • Step 2: Concavity f′ (x) = 6x2 − 6x − 12 =⇒ f′′ (x) = 12x − 6 = 6(2x − 1) Another sign chart: . . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 13 / 47
    • Step 2: Concavity f′ (x) = 6x2 − 6x − 12 =⇒ f′′ (x) = 12x − 6 = 6(2x − 1) Another sign chart: . .′′ (x) f . ./2 1 f .(x) . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 13 / 47
    • Step 2: Concavity f′ (x) = 6x2 − 6x − 12 =⇒ f′′ (x) = 12x − 6 = 6(2x − 1) Another sign chart: . − . − .′′ (x) f . ./2 1 f .(x) . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 13 / 47
    • Step 2: Concavity f′ (x) = 6x2 − 6x − 12 =⇒ f′′ (x) = 12x − 6 = 6(2x − 1) Another sign chart: . − . − . + + .′′ (x) f . ./2 1 f .(x) . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 13 / 47
    • Step 2: Concavity f′ (x) = 6x2 − 6x − 12 =⇒ f′′ (x) = 12x − 6 = 6(2x − 1) Another sign chart: . − . − . + + .′′ (x) f . . ⌢ ./2 1 f .(x) . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 13 / 47
    • Step 2: Concavity f′ (x) = 6x2 − 6x − 12 =⇒ f′′ (x) = 12x − 6 = 6(2x − 1) Another sign chart: . − . − . + + .′′ (x) f . . ⌢ ./2 1 . ⌣ f .(x) . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 13 / 47
    • Step 2: Concavity f′ (x) = 6x2 − 6x − 12 =⇒ f′′ (x) = 12x − 6 = 6(2x − 1) Another sign chart: . − . − . + + .′′ (x) f . . ⌢ ./2 1 . ⌣ f .(x) I .P . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 13 / 47
    • Step 3: One sign chart to rule them all Remember, f(x) = 2x3 − 3x2 − 12x. . . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 14 / 47
    • Step 3: One sign chart to rule them all Remember, f(x) = 2x3 − 3x2 − 12x. − . . . . + − . . + .′ (x) f . ↗− ↘ . . 1 . ↘ . 2 . ↗ . m . onotonicity . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 14 / 47
    • Step 3: One sign chart to rule them all Remember, f(x) = 2x3 − 3x2 − 12x. . . + − . . − . . + .′ (x) f . ↗− . . 1 ↘ . ↘ . . 2 ↗ . m .′′ onotonicity − . − − . − . . + + . + + f . (x) . ⌢ . ⌢ 1/2 . . ⌣ . ⌣ c . oncavity . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 14 / 47
    • Step 3: One sign chart to rule them all Remember, f(x) = 2x3 − 3x2 − 12x. . . + − . . − . . + .′ (x) f . ↗− . . 1 ↘ . ↘ . . 2 ↗ . m .′′ onotonicity − . − − . − . . + + . + + f . (x) . ⌢ ⌢ ./2 . . 1 ⌣ . ⌣ c . oncavity 7 .. − . 6 1/2 −. . 20 f .(x) . − . 1 . 1/2 2 . s . hape of f m . ax I .P m . in . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 14 / 47
    • Combinations of monotonicity and concavity I .I I . . I .II I .V . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 15 / 47
    • Combinations of monotonicity and concavity . decreasing, concave down I .I I . . I .II I .V . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 15 / 47
    • Combinations of monotonicity and concavity . . increasing, decreasing, concave concave down down I .I I . . I .II I .V . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 15 / 47
    • Combinations of monotonicity and concavity . . increasing, decreasing, concave concave down down I .I I . . I .II I .V . decreasing, concave up . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 15 / 47
    • Combinations of monotonicity and concavity . . increasing, decreasing, concave concave down down I .I I . . I .II I .V . . decreasing, increasing, concave up concave up . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 15 / 47
    • Step 3: One sign chart to rule them all Remember, f(x) = 2x3 − 3x2 − 12x. . . + − . . − . . + .′ (x) f . ↗− . . 1 ↘ . ↘ . . 2 ↗ . m .′′ onotonicity − . − − . − . . + + . + + f . (x) . ⌢ ⌢ ./2 . . 1 ⌣ . ⌣ c . oncavity 7 .. − . 6 1/2 −. . 20 f .(x) . . . 1 − . 1/2 2 . s . hape of f m . ax I .P m . in . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 16 / 47
    • Step 3: One sign chart to rule them all Remember, f(x) = 2x3 − 3x2 − 12x. . . + − . . − . . + .′ (x) f . ↗− . . 1 ↘ . ↘ . . 2 ↗ . m .′′ onotonicity − . − − . − . . + + . + + f . (x) . ⌢ ⌢ ./2 . . 1 ⌣ . ⌣ c . oncavity 7 .. − . 6 1/2 −. . 20 f .(x) . . . 1 . ./2 − 1 2 . s . hape of f m . ax I .P m . in . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 16 / 47
    • Step 3: One sign chart to rule them all Remember, f(x) = 2x3 − 3x2 − 12x. . . + − . . − . . + .′ (x) f . ↗− . . 1 ↘ . ↘ . . 2 ↗ . m .′′ onotonicity − . − − . − . . + + . + + f . (x) . ⌢ ⌢ ./2 . . 1 ⌣ . ⌣ c . oncavity 7 .. − . 6 1/2 −. . 20 f .(x) . . . 1 . ./2 . − 1 2 . s . hape of f m . ax I .P m . in . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 16 / 47
    • Step 3: One sign chart to rule them all Remember, f(x) = 2x3 − 3x2 − 12x. . . + − . . − . . + .′ (x) f . ↗− . . 1 ↘ . ↘ . . 2 ↗ . m .′′ onotonicity − . − − . − . . + + . + + f . (x) . ⌢ ⌢ ./2 . . 1 ⌣ . ⌣ c . oncavity 7 .. − . 6 1/2 −. . 20 f .(x) . . . 1 . ./2 . − 1 2 . . s . hape of f m . ax I .P m . in . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 16 / 47
    • Step 4: Graph f .(x) .(x) = 2x3 − 3x2 − 12x f ( √ ) . −1, 7) ( . . 3− 4105 , 0 . 0, 0) ( . . . . 1/2, −61/2) ( ( . x √ ) . . 3+ 4105 , 0 . 2, −20) ( . 7 .. − . 61/2 −. . 20 f .(x) . . . 1 . ./2 . − 1 2 . . s . hape of f m . ax I .P m . in . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 17 / 47
    • Step 4: Graph f .(x) .(x) = 2x3 − 3x2 − 12x f ( √ ) . −1, 7) ( . . 3− 4105 , 0 . 0, 0) ( . . . . 1/2, −61/2) ( ( . x √ ) . . 3+ 4105 , 0 . 2, −20) ( . 7 .. − . 61/2 −. . 20 f .(x) . . . 1 . ./2 . − 1 2 . . s . hape of f m . ax I .P m . in . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 17 / 47
    • Step 4: Graph f .(x) .(x) = 2x3 − 3x2 − 12x f ( √ ) . −1, 7) ( . . 3− 4105 , 0 . 0, 0) ( . . . . 1/2, −61/2) ( ( . x √ ) . . 3+ 4105 , 0 . 2, −20) ( . 7 .. − . 61/2 −. . 20 f .(x) . . . 1 . ./2 . − 1 2 . . s . hape of f m . ax I .P m . in . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 17 / 47
    • Step 4: Graph f .(x) .(x) = 2x3 − 3x2 − 12x f ( √ ) . −1, 7) ( . . 3− 4105 , 0 . 0, 0) ( . . . . 1/2, −61/2) ( ( . x √ ) . . 3+ 4105 , 0 . 2, −20) ( . 7 .. − . 61/2 −. . 20 f .(x) . . . 1 . ./2 . − 1 2 . . s . hape of f m . ax I .P m . in . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 17 / 47
    • Step 4: Graph f .(x) .(x) = 2x3 − 3x2 − 12x f ( √ ) . −1, 7) ( . . 3− 4105 , 0 . 0, 0) ( . . . . 1/2, −61/2) ( ( . x √ ) . . 3+ 4105 , 0 . 2, −20) ( . 7 .. − . 61/2 −. . 20 f .(x) . . . 1 . ./2 . − 1 2 . . s . hape of f m . ax I .P m . in . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 17 / 47
    • Graphing a quartic Example Graph f(x) = x4 − 4x3 + 10 . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 18 / 47
    • Graphing a quartic Example Graph f(x) = x4 − 4x3 + 10 (Step 0) We know f(0) = 10 and lim f(x) = +∞. Not too many other x→±∞ points on the graph are evident. . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 18 / 47
    • Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 19 / 47
    • Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. . . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 19 / 47
    • Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. 0 .. . x2 4 0 . . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 19 / 47
    • Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. . .. + 0 . x2 4 0 . . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 19 / 47
    • Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. . .. + 0 . + . x2 4 0 . . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 19 / 47
    • Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. . .. + 0 . + . + . x2 4 0 . . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 19 / 47
    • Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. . .. + 0 . + . + . x2 4 0 . 0 .. . x − 3) ( 3 . . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 19 / 47
    • Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. . .. + 0 . + . + . x2 4 0 . − . 0 .. . x − 3) ( 3 . . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 19 / 47
    • Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. . .. + 0 . + . + . x2 4 0 . − . − . 0 .. . x − 3) ( 3 . . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 19 / 47
    • Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. . .. + 0 . + . + . x2 4 0 . − . − . .. . 0 + . x − 3) ( 3 . . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 19 / 47
    • Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. . .. + 0 . + . + . x2 4 0 . − . − . .. . 0 + . x − 3) ( 3 . 0 .. 0 .. .′ (x) f 0 . 3 . f .(x) . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 19 / 47
    • Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. . .. + 0 . + . + . x2 4 0 . − . − . .. . 0 + . x − 3) ( 3 . − 0 . .. 0 .. .′ (x) f 0 . 3 . f .(x) . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 19 / 47
    • Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. . .. + 0 . + . + . x2 4 0 . − . − . .. . 0 + . x − 3) ( 3 . − 0 . .. − . 0 .. .′ (x) f 0 . 3 . f .(x) . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 19 / 47
    • Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. . .. + 0 . + . + . x2 4 0 . − . − . .. . 0 + . x − 3) ( 3 . − 0 . .. − . .. . 0 + .′ (x) f 0 . 3 . f .(x) . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 19 / 47
    • Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. . .. + 0 . + . + . x2 4 0 . − . − . .. . 0 + . x − 3) ( 3 . − 0 . .. − . .. . 0 + .′ (x) f ↘ 0 . . 3 . f .(x) . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 19 / 47
    • Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. . .. + 0 . + . + . x2 4 0 . − . − . .. . 0 + . x − 3) ( 3 . − 0 . .. − . .. . 0 + .′ (x) f ↘ 0 . . ↘ . 3 . f .(x) . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 19 / 47
    • Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. . .. + 0 . + . + . x2 4 0 . − . − . .. . 0 + . x − 3) ( 3 . − 0 . .. − . .. . 0 + .′ (x) f ↘ 0 . . ↘ . 3 ↗ . . f .(x) . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 19 / 47
    • Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. . .. + 0 . + . + . x2 4 0 . − . − . .. . 0 + . x − 3) ( 3 . − 0 . .. − . .. . 0 + .′ (x) f ↘ 0 . . ↘ . 3 ↗ . . f .(x) m . in . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 19 / 47
    • 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 . .. . + 1 . 2x 0 . . . . . . .
    • Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: − 0 . .. . + . + 1 . 2x 0 . . . . . . .
    • Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: − 0 . .. . + . + 1 . 2x 0 . . . . . . .
    • Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: − 0 . .. . + . + 1 . 2x 0 . . . . . . .
    • Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: − 0 . .. . + . + 1 . 2x 0 . − . − . 0 .. . −2 x 2 . . . . . . .
    • Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: − 0 . .. . + . + 1 . 2x 0 . − . − . 0 .. . + . −2 x 2 . . . . . . .
    • Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: − 0 . .. . + . + 1 . 2x 0 . − . − . 0 .. . + . −2 x 2 . . . . . . .
    • Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: − 0 . .. . + . + 1 . 2x 0 . − . − . 0 .. . + . −2 x 2 . . . . . . .
    • Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: − 0 . .. . + . + 1 . 2x 0 . − . − . 0 .. . + . −2 x 2 . . + .. + 0 − . − 0 .. .′′ (x) f 0 . 2 . f .(x) . . . . . .
    • Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: − 0 . .. . + . + 1 . 2x 0 . − . − . 0 .. . + . −2 x 2 . . + .. + 0 − . − 0 .. . + + .′′ (x) f 0 . 2 . f .(x) . . . . . .
    • Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: − 0 . .. . + . + 1 . 2x 0 . − . − . 0 .. . + . −2 x 2 . . + .. + 0 − . − 0 .. . + + .′′ (x) f . . ⌣ 0 2 . f .(x) . . . . . .
    • Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: − 0 . .. . + . + 1 . 2x 0 . − . − . 0 .. . + . −2 x 2 . . + .. + 0 − . − 0 .. . + + .′′ (x) f . . ⌣ 0 . ⌢ 2 . f .(x) . . . . . .
    • Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: − 0 . .. . + . + 1 . 2x 0 . − . − . 0 .. . + . −2 x 2 . . + .. + 0 − . − 0 .. . + + .′′ (x) f . . ⌣ 0 . ⌢ 2 . . ⌣ f .(x) . . . . . .
    • Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: − 0 . .. . + . + 1 . 2x 0 . − . − . 0 .. . + . −2 x 2 . . + .. + 0 − . − 0 .. . + + .′′ (x) f . . ⌣ 0 . ⌢ 2 . . ⌣ f .(x) I .P . . . . . .
    • Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: − 0 . .. . + . + 1 . 2x 0 . − . − . 0 .. . + . −2 x 2 . . + .. + 0 − . − 0 .. . + + .′′ (x) f . . ⌣ 0 . ⌢ 2 . . ⌣ f .(x) I .P I .P . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 20 / 47
    • Step 3: Grand Unified Sign Chart . Remember, f(x) = x4 − 4x3 + 10. − 0 . .. − . − 0 + . .. . .′ (x) f ↘ 0 . . ↘ . ↘ 3 ↗ . . . m .′′ onotonicity . + .. + 0 − . − .. . + . + 0+ + f . (x) . . ⌣ 0 . ⌢ 2 . . ⌣ . ⌣ c . oncavity 1. .0 − −. . .6 . 17 f .(x) 0 . 2 . 3 . s . hape I .P I .P m . in . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 21 / 47
    • Step 3: Grand Unified Sign Chart . Remember, f(x) = x4 − 4x3 + 10. − 0 . .. − . − 0 + . .. . .′ (x) f ↘ 0 . . ↘ . ↘ 3 ↗ . . . m .′′ onotonicity . + .. + 0 − . − .. . + . + 0+ + f . (x) . . ⌣ 0 . ⌢ 2 . . ⌣ . ⌣ c . oncavity 1. .0 − −. . .6 . 17 f .(x) . .0 2 . 3 . s . hape I .P I .P m . in . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 21 / 47
    • Step 3: Grand Unified Sign Chart . Remember, f(x) = x4 − 4x3 + 10. − 0 . .. − . − 0 + . .. . .′ (x) f ↘ 0 . . ↘ . ↘ 3 ↗ . . . m .′′ onotonicity . + .. + 0 − . − .. . + . + 0+ + f . (x) . . ⌣ 0 . ⌢ 2 . . ⌣ . ⌣ c . oncavity 1. .0 − −. . .6 . 17 f .(x) . .0 . 2 . 3 . s . hape I .P I .P m . in . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 21 / 47
    • Step 3: Grand Unified Sign Chart . Remember, f(x) = x4 − 4x3 + 10. − 0 . .. − . − 0 + . .. . .′ (x) f ↘ 0 . . ↘ . ↘ 3 ↗ . . . m .′′ onotonicity . + .. + 0 − . − .. . + . + 0+ + f . (x) . . ⌣ 0 . ⌢ 2 . . ⌣ . ⌣ c . oncavity 1. .0 − −. . .6 . 17 f .(x) . .0 . 2 . . . 3 s . hape I .P I .P m . in . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 21 / 47
    • Step 3: Grand Unified Sign Chart . Remember, f(x) = x4 − 4x3 + 10. − 0 . .. − . − 0 + . .. . .′ (x) f ↘ 0 . . ↘ . ↘ 3 ↗ . . . m .′′ onotonicity . + .. + 0 − . − .. . + . + 0+ + f . (x) . . ⌣ 0 . ⌢ 2 . . ⌣ . ⌣ c . oncavity 1. .0 − −. . .6 . 17 f .(x) . .0 . 2 . . . . 3 s . hape I .P I .P m . in . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 21 / 47
    • Step 4: Graph y . .(x) = x4 − 4x3 + 10 f . 0, 10) ( . . . x . . . 2, −6) ( . 3, −17) ( 1. .0 − −. . .6 . 17 f .(x) . .0 . 2 . . . . 3 s . hape I .P I .P . in m . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 22 / 47
    • Step 4: Graph y . .(x) = x4 − 4x3 + 10 f . 0, 10) ( . . . x . . . 2, −6) ( . 3, −17) ( 1. .0 − −. . .6 . 17 f .(x) . .0 . 2 . . . . 3 s . hape I .P I .P . in m . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 22 / 47
    • Step 4: Graph y . .(x) = x4 − 4x3 + 10 f . 0, 10) ( . . . x . . . 2, −6) ( . 3, −17) ( 1. .0 − −. . .6 . 17 f .(x) . .0 . 2 . . . . 3 s . hape I .P I .P . in m . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 22 / 47
    • Step 4: Graph y . .(x) = x4 − 4x3 + 10 f . 0, 10) ( . . . x . . . 2, −6) ( . 3, −17) ( 1. .0 − −. . .6 . 17 f .(x) . .0 . 2 . . . . 3 s . hape I .P I .P . in m . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 22 / 47
    • Step 4: Graph y . .(x) = x4 − 4x3 + 10 f . 0, 10) ( . . . x . . . 2, −6) ( . 3, −17) ( 1. .0 − −. . .6 . 17 f .(x) . .0 . 2 . . . . 3 s . hape I .P I .P . in m . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 22 / 47
    • Outline The Procedure Simple examples A cubic function A quartic function More Examples Points of nondifferentiability Horizontal asymptotes Vertical asymptotes Trigonometric and polynomial together Logarithmic . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 23 / 47
    • Example √ Graph f(x) = x + |x| . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 24 / 47
    • Example √ Graph f(x) = x + |x| This function looks strange because of the absolute value. But whenever we become nervous, we can just take cases. . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 24 / 47
    • 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 positive. . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 25 / 47
    • 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 positive. Are there negative numbers which are zeroes for f? . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 25 / 47
    • 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 positive. Are there negative numbers which are zeroes for f? √ x + −x = 0 √ −x = −x −x = x2 x2 + x = 0 The only solutions are x = 0 and x = −1 . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 25 / 47
    • Step 0: Asymptotic behavior √ f(x) = x + |x| lim f(x) = ∞, because both terms tend to ∞. x→∞ . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 26 / 47
    • 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 as x→−∞ √ lim (−y + y) y→+∞ . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 26 / 47
    • 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 as x→−∞ √ lim (−y + y) y→+∞ √ √ √ y+y lim (−y + y) = lim ( y − y) · √ y→+∞ y→∞ y+y y − y2 = lim √ = −∞ y→∞ y+y . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 26 / 47
    • 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 . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 27 / 47
    • 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 Notice f′ (x) > 0 when x > 0 (so no critical points here) . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 27 / 47
    • 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 Notice f′ (x) > 0 when x > 0 (so no critical points here) lim f′ (x) = ∞ (so 0 is a critical point) x→0+ . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 27 / 47
    • 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 Notice f′ (x) > 0 when x > 0 (so no critical points here) lim f′ (x) = ∞ (so 0 is a critical point) x→0+ lim f′ (x) = 1 (so the graph is asymptotic to a line of slope 1) x→∞ . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 27 / 47
    • Step 1: The derivative √ Remember, f(x) = x + |x|. If x is negative, we have d ( √ ) 1 f′ (x) = x + −x = 1 − √ dx 2 −x Notice lim f′ (x) = −∞ (other side of the critical point) x→0− . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 28 / 47
    • Step 1: The derivative √ Remember, f(x) = x + |x|. If x is negative, we have d ( √ ) 1 f′ (x) = x + −x = 1 − √ dx 2 −x Notice lim f′ (x) = −∞ (other side of the critical point) x→0− lim f′ (x) = 1 (asymptotic to a line of slope 1) x→−∞ . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 28 / 47
    • Step 1: The derivative √ Remember, f(x) = x + |x|. If x is negative, we have d ( √ ) 1 f′ (x) = x + −x = 1 − √ dx 2 −x Notice lim f′ (x) = −∞ (other side of the critical point) x→0− lim f′ (x) = 1 (asymptotic 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 . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 28 / 47
    • Step 1: Monotonicity  1 1 + √  if x > 0 f′ (x) = 2 x 1 − √  1 if x < 0 2 −x We can’t make a multi-factor sign chart because of the absolute value, but we can test points in between critical points. 0 .. ∓. . ∞ .′ (x) f −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 multi-factor sign chart because of the absolute value, but we can test points in between critical points. . + 0 .. ∓. . ∞ .′ (x) f −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 multi-factor sign chart because of the absolute value, but we can test points in between critical points. . + 0 −∓ . .. . . ∞ .′ (x) f −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 multi-factor sign chart because of the absolute value, but we can test points in between critical points. . + 0 −∓ . .. . . ∞ . + .′ (x) f −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 multi-factor sign chart because of the absolute value, but we can test points in between critical points. . + 0 −∓ . .. . . ∞ . + .′ (x) f ↗ . −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 multi-factor sign chart because of the absolute value, but we can test points in between critical points. . + 0 −∓ . .. . . ∞ . + .′ (x) f ↗ . −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 multi-factor sign chart because of the absolute value, but we can test points in between critical points. . + 0 −∓ . .. . . ∞ . + .′ (x) f ↗ . −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 multi-factor sign chart because of the absolute value, but we can test points in between critical points. . + 0 −∓ . .. . . ∞ . + .′ (x) f ↗ . −4 ↘ 0 . 1. . ↗ . 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 multi-factor sign chart because of the absolute value, but we can test points in between critical points. . + 0 −∓ . .. . . ∞ . + .′ (x) f ↗ . −4 ↘ 0 . .1 . . ↗ . f .(x) . max min . . . . . .
    • Step 1: Monotonicity  1 1 + √  if x > 0 f′ (x) = 2 x 1 − √  1 if x < 0 2 −x We can’t make a multi-factor sign chart because of the absolute value, but we can test points in between critical points. . + 0 −∓ . .. . . ∞ . + .′ (x) f ↗ . −4 ↘ 0 . .1 . . ↗ . f .(x) . max min . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 29 / 47
    • Step 2: Concavity If x > 0, then ( ) d 1 1 f′′ (x) = 1 + x−1/2 = − x−3/2 dx 2 4 This is negative whenever x > 0. . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 30 / 47
    • Step 2: Concavity If x > 0, then ( ) d 1 1 f′′ (x) = 1 + x−1/2 = − x−3/2 dx 2 4 This is negative whenever x > 0. If x < 0, then ( ) ′′ d 1 −1/2 1 f (x) = 1 − (−x) = − (−x)−3/2 dx 2 4 which is also always negative for negative x. . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 30 / 47
    • Step 2: Concavity If x > 0, then ( ) d 1 1 f′′ (x) = 1 + x−1/2 = − x−3/2 dx 2 4 This is negative whenever x > 0. If x < 0, then ( ) ′′ d 1 −1/2 1 f (x) = 1 − (−x) = − (−x)−3/2 dx 2 4 which is also always negative for negative x. 1 In other words, f′′ (x) = − |x|−3/2 . 4 . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 30 / 47
    • Step 2: Concavity If x > 0, then ( ) d 1 1 f′′ (x) = 1 + x−1/2 = − x−3/2 dx 2 4 This is negative whenever x > 0. If x < 0, then ( ) ′′ d 1 −1/2 1 f (x) = 1 − (−x) = − (−x)−3/2 dx 2 4 which is also always negative for negative x. 1 In other words, f′′ (x) = − |x|−3/2 . 4 Here is the sign chart: ′′ − . − −. . ∞ − . − . . (x) f . ⌢ . ⌢ . 0 . f .(x) . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 30 / 47
    • Step 3: Synthesis Now we can put these things together. √ f(x) = x + |x| ′ . 1 + . + 0 −∓ . .. . . ∞ . + +. f . 1 (x) ↗ . ↗ . −1 ↘ 0 . 4. . ↗ . ↗m . . onotonicity − . ∞ − . − − . . . − ∞ − − . − − . f′′ . ∞ (x) . ⌢ . ⌢ . . ⌢ 0 . ⌢ . . oncavity ⌢c − . ∞ 0 .. .1 4 0 .. . ∞ + .(x) f . − . 1 −1 . .4 0 . s . hape . . zero max min . . . . . .
    • Step 3: Synthesis Now we can put these things together. √ f(x) = x + |x| ′ . 1 + . + 0 −∓ . .. . . ∞ . + +. f . 1 (x) ↗ . ↗ . −1 ↘ 0 . 4. . ↗ . ↗m . . onotonicity − . ∞ − . − − . . . − ∞ − − . − − . f′′ . ∞ (x) . ⌢ . ⌢ . . ⌢ 0 . ⌢ . . oncavity ⌢c − . ∞ 0 .. .1 4 0 .. . ∞ + .(x) f . . . 1 − −1 . .4 0 . s . hape . . zero max min . . . . . .
    • Step 3: Synthesis Now we can put these things together. √ f(x) = x + |x| ′ . 1 + . + 0 −∓ . .. . . ∞ . + +. f . 1 (x) ↗ . ↗ . −1 ↘ 0 . 4. . ↗ . ↗m . . onotonicity − . ∞ − . − − . . . − ∞ − − . − − . f′′ . ∞ (x) . ⌢ . ⌢ . . ⌢ 0 . ⌢ . . oncavity ⌢c − . ∞ 0 .. .1 4 0 .. . ∞ + .(x) f . . . 1 − . −1 . .4 0 . s . hape . . zero max min . . . . . .
    • Step 3: Synthesis Now we can put these things together. √ f(x) = x + |x| ′ . 1 + . + 0 −∓ . .. . . ∞ . + +. f . 1 (x) ↗ . ↗ . −1 ↘ 0 . 4. . ↗ . ↗m . . onotonicity − . ∞ − . − − . . . − ∞ − − . − − . f′′ . ∞ (x) . ⌢ . ⌢ . . ⌢ 0 . ⌢ . . oncavity ⌢c − . ∞ 0 .. .1 4 0 .. . ∞ + .(x) f . . . 1 − . −1 . .4 . . 0 s . hape . . zero max min . . . . . .
    • Step 3: Synthesis Now we can put these things together. √ f(x) = x + |x| ′ . 1 + . + 0 −∓ . .. . . ∞ . + +. f . 1 (x) ↗ . ↗ . −1 ↘ 0 . 4. . ↗ . ↗m . . onotonicity − . ∞ − . − − . . . − ∞ − − . − − . f′′ . ∞ (x) . ⌢ . ⌢ . . ⌢ 0 . ⌢ . . oncavity ⌢c − . ∞ 0 .. .1 4 0 .. . ∞ + .(x) f . . . 1 − . −1 . .4 . . 0 . s . hape . . zero max min . . . . . .
    • Step 3: Synthesis Now we can put these things together. √ f(x) = x + |x| ′ . 1 + . + 0 −∓ . .. . . ∞ . + +. f . 1 (x) ↗ . ↗ . −1 ↘ 0 . 4. . ↗ . ↗m . . onotonicity − . ∞ − . − − . . . − ∞ − − . − − . f′′ . ∞ (x) . ⌢ . ⌢ . . ⌢ 0 . ⌢ . . oncavity ⌢c − . ∞ 0 .. .1 4 0 .. . ∞ + .(x) f . . . 1 − . −1 . .4 . . 0 . s . hape . . zero max min . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 31 / 47
    • Graph √ f(x) = x + |x| f .(x) .−1, 1) ( 4 4 . −1, 0) ( . . . x . . 0, 0) ( − 0 . ∞ .. .1 4 0 .. . ∞ .(x) + f . . − . 1 . . .1 . . −4 0 . s . hape . . zero max min . . . . . .
    • Graph √ f(x) = x + |x| f .(x) .−1, 1) ( 4 4 . −1, 0) ( . . . x . . 0, 0) ( − 0 . ∞ .. .1 4 0 .. . ∞ .(x) + f . . − . 1 . . .1 . . −4 0 . s . hape . . zero max min . . . . . .
    • Graph √ f(x) = x + |x| f .(x) .−1, 1) ( 4 4 . −1, 0) ( . . . x . . 0, 0) ( − 0 . ∞ .. .1 4 0 .. . ∞ .(x) + f . . − . 1 . . .1 . . −4 0 . s . hape . . zero max min . . . . . .
    • Graph √ f(x) = x + |x| f .(x) .−1, 1) ( 4 4 . −1, 0) ( . . . x . . 0, 0) ( − 0 . ∞ .. .1 4 0 .. . ∞ .(x) + f . . − . 1 . . .1 . . −4 0 . s . hape . . zero max min . . . . . .
    • Graph √ f(x) = x + |x| f .(x) .−1, 1) ( 4 4 . −1, 0) ( . . . x . . 0, 0) ( − 0 . ∞ .. .1 4 0 .. . ∞ .(x) + f . . − . 1 . . .1 . . −4 0 . s . hape . . zero max min . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 32 / 47
    • Example Graph f(x) = xe−x 2 . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 33 / 47
    • Example Graph f(x) = xe−x 2 Before taking derivatives, we notice that f is odd, that f(0) = 0, and lim f(x) = 0 x→∞ . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 33 / 47
    • Step 1: Monotonicity If f(x) = xe−x , then 2 ( ) f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x 2 2 2 ( √ )( √ ) = 1 − 2x 1 + 2x e−x 2 The factor e−x is always positive so it doesn’t figure into the sign of 2 f′ (x). So our sign chart looks like this: . + .. + 0 . − . √ √. . − 1 2x . 1/2 − . 0 .. . + . + √ √ 1 . + 2x − . 1/2 − . 0 .. . + 0 . − . .′ (x) f √ √. ↘ . − 1/2 ↗ . ↘ . f .(x) . . . . 1/2 min max . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 34 / 47
    • Step 2: Concavity If f′ (x) = (1 − 2x2 )e−x , we know 2 ( ) f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 2 = 2x(2x2 − 3)e−x 2 − . − . 0 .. . + . + 2 .x 0 . − . − . − . 0 . . + √ √ √. . 2x − 3 . 3/2 − . 0 .. . + . + . + √ √ √ . 2x + 3 − . 3/2 − . − .. . + 0 + 0 .. − . − .. 0 . + + .′′ (x) f . ⌢ √ . ⌣ ⌢ √3 . . ⌣ − 3/2 . 0 . f .(x) . . . . /2 IP IP IP . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 35 / 47
    • Step 3: Synthesis f(x) = xe−x 2 − . − 0 + . .. . + . − − .′ (x) f √ . . √. . 0 . ↘ . ↘ ↗ . . 1/2 . − ↗ ↘ . . 1/2. ↘ . m . onotonicity − . − .. . + 0+ + 0 − . + .. . − − 0 . − .. . + + .′′ (x) f . ⌢ √. ⌣ ⌣ . . . . √3 . − . 3/2 0 ⌢ ⌢ . /2 ⌣ c . oncavity √ √ − . 3 2e3 − . √1 .√1 . 33 f .(x) . . 2e 0 .. . √2e 2e . √ . √ √ . . − . . . . − 1/2 . 3/2 . . . 0 . . 1/2 . 3/2 . . s . hape IP min IP max IP . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 36 / 47
    • Step 4: Graph f .(x) (√ )( . 1/2, √1 √ √ ) 2e . 3 .(x) = xe−x 2 3/2, f . 2e3 . . x . . 0, 0) ( ( . √ √ ) . . − 3/2, − 2e3 ( √ 3 ) . − 1/2, − √1 2e √ √ − 2e3 √1 . 3 − . .√1 . 2e33 f .(x) . √2e 0 . 2e . √ . . . . . √ . . √. . − 3 . 1/2 . . − . 0 s . hape . . . /2 . .1/2 . 3/2 IP min IP max IP . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 37 / 47
    • Example 1 1 Graph f(x) = + 2 x x . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 38 / 47
    • Step 0 Find when f is positive, negative, zero, not defined. . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 39 / 47
    • Step 0 Find when f is positive, negative, zero, not defined. We need to factor f: 1 1 x+1 f(x) = + = . x x2 x2 This means f is 0 at −1 and has trouble at 0. In fact, x+1 lim = ∞, x→0 x2 so x = 0 is a vertical asymptote of the graph. . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 39 / 47
    • Step 0 Find when f is positive, negative, zero, not defined. We need to factor f: 1 1 x+1 f(x) = + = . x x2 x2 This means f is 0 at −1 and has trouble at 0. In fact, x+1 lim = ∞, x→0 x2 so x = 0 is a vertical asymptote of the graph. We can make a sign chart as follows: − . 0 .. . . + x . +1 − . 1 . + 0 .. . + .2 x 0 . − . .. . 0 + ∞ .. . + f .(x) − . 1 0 . . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 39 / 47
    • Step 0, continued For horizontal asymptotes, notice that x+1 lim = 0, x→∞ x2 so y = 0 is a horizontal asymptote of the graph. The same is true at −∞. . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 40 / 47
    • Step 1: Monotonicity . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 41 / 47
    • Step 1: Monotonicity We have 1 2 x+2 f′ (x) = − − =− 3 . x2 x3 x The critical points are x = −2 and x = 0. We have the following sign chart: . + 0 .. . − . − . (x + 2) − . 2 − . 0 .. . + .3 x 0 . . . . . . .
    • Step 1: Monotonicity We have 1 2 x+2 f′ (x) = − − =− 3 . x2 x3 x The critical points are x = −2 and x = 0. We have the following sign chart: . + 0 .. . − . − . (x + 2) − . 2 − . 0 .. . + .3 x 0 . − . 0 .. . + ∞ .. − . .′ (x) f − . 2 0 . f .(x) . . . . . .
    • Step 1: Monotonicity We have 1 2 x+2 f′ (x) = − − =− 3 . x2 x3 x The critical points are x = −2 and x = 0. We have the following sign chart: . + 0 .. . − . − . (x + 2) − . 2 − . 0 .. . + .3 x 0 . − .. . 0 . + ∞ .. − . .′ (x) f ↘ . 2 . − 0 . f .(x) . . . . . .
    • Step 1: Monotonicity We have 1 2 x+2 f′ (x) = − − =− 3 . x2 x3 x The critical points are x = −2 and x = 0. We have the following sign chart: . + 0 .. . − . − . (x + 2) − . 2 − . 0 .. . + .3 x 0 . − .. . 0 . + ∞ .. − . .′ (x) f ↘ . 2 . − ↗ . 0 . f .(x) . . . . . .
    • Step 1: Monotonicity We have 1 2 x+2 f′ (x) = − − =− 3 . x2 x3 x The critical points are x = −2 and x = 0. We have the following sign chart: . + 0 .. . − . − . (x + 2) − . 2 − . .. . 0 + .3 x 0 . − .. . 0 . + ∞ − .. . .′ (x) f ↘ . 2 . − ↗ . 0 ↘ . . f .(x) . . . . . .
    • Step 1: Monotonicity We have 1 2 x+2 f′ (x) = − − =− 3 . x2 x3 x The critical points are x = −2 and x = 0. We have the following sign chart: . + 0 .. . − . − . (x + 2) − . 2 − . .. . 0 + .3 x 0 . − .. . 0 . + ∞ − .. . .′ (x) f ↘ . 2 . − ↗ . 0 ↘ . . f .(x) m . in . . . . . .
    • Step 1: Monotonicity We have 1 2 x+2 f′ (x) = − − =− 3 . x2 x3 x The critical points are x = −2 and x = 0. We have the following sign chart: . + 0 .. . − . − . (x + 2) − . 2 − . .. . 0 + .3 x 0 . − .. . 0 . + ∞ − .. . .′ (x) f ↘ . 2 . − ↗ . 0 ↘ . . f .(x) m . in V .A . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 41 / 47
    • Step 2: Concavity . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 42 / 47
    • Step 2: Concavity We have 2 6 2(x + 3) f′′ (x) = + = . x3 x4 x4 The critical points of f′ are −3 and 0. Sign chart: − . 0 .. . . + . x + 3) ( − . 3 . + 0 .. . + .4 x 0 . 0 .. ∞ .. .′′ (x) f − . 3 0 . f .(x) . . . . . .
    • Step 2: Concavity We have 2 6 2(x + 3) f′′ (x) = + = . x3 x4 x4 The critical points of f′ are −3 and 0. Sign chart: − . 0 .. . . + . x + 3) ( − . 3 . + 0 .. . + .4 x 0 . − . − .. 0 ∞ .. .′′ (x) f − . 3 0 . f .(x) . . . . . .
    • Step 2: Concavity We have 2 6 2(x + 3) f′′ (x) = + = . x3 x4 x4 The critical points of f′ are −3 and 0. Sign chart: − . 0 .. . . + . x + 3) ( − . 3 . + 0 .. . + .4 x 0 . − . − .. 0 . + + ∞ .. .′′ (x) f − . 3 0 . f .(x) . . . . . .
    • Step 2: Concavity We have 2 6 2(x + 3) f′′ (x) = + = . x3 x4 x4 The critical points of f′ are −3 and 0. Sign chart: − . 0 .. . . + . x + 3) ( − . 3 . + .. . 0 + .4 x 0 . − . − .. 0 . + + ∞ + .. . + .′′ (x) f − . 3 0 . f .(x) . . . . . .
    • Step 2: Concavity We have 2 6 2(x + 3) f′′ (x) = + = . x3 x4 x4 The critical points of f′ are −3 and 0. Sign chart: − . 0 .. . . + . x + 3) ( − . 3 . + .. . 0 + .4 x 0 . − . − .. 0 . + + ∞ + .. . + .′′ (x) f . ⌢ . 3 − 0 . f .(x) . . . . . .
    • Step 2: Concavity We have 2 6 2(x + 3) f′′ (x) = + = . x3 x4 x4 The critical points of f′ are −3 and 0. Sign chart: − . 0 .. . . + . x + 3) ( − . 3 . + .. . 0 + .4 x 0 . − . − .. 0 . + + ∞ + .. . + .′′ (x) f . ⌢ . 3 − . ⌣ 0 . f .(x) . . . . . .
    • Step 2: Concavity We have 2 6 2(x + 3) f′′ (x) = + = . x3 x4 x4 The critical points of f′ are −3 and 0. Sign chart: − . 0 .. . . + . x + 3) ( − . 3 . + .. . 0 + .4 x 0 . − . − .. 0 . + + ∞ + .. . + .′′ (x) f . ⌢ . 3 − . ⌣ . . 0 ⌣ f .(x) . . . . . .
    • Step 2: Concavity We have 2 6 2(x + 3) f′′ (x) = + = . x3 x4 x4 The critical points of f′ are −3 and 0. Sign chart: − . 0 .. . . + . x + 3) ( − . 3 . + .. . 0 + .4 x 0 . − . − .. 0 . + + ∞ + .. . + .′′ (x) f . ⌢ . 3 − . ⌣ . . 0 ⌣ f .(x) I .P . . . . . .
    • Step 2: Concavity We have 2 6 2(x + 3) f′′ (x) = + = . x3 x4 x4 The critical points of f′ are −3 and 0. Sign chart: − . 0 .. . . + . x + 3) ( − . 3 . + .. . 0 + .4 x 0 . − . − .. 0 . + + ∞ + .. . + .′′ (x) f . ⌢ . 3 − . ⌣ . . 0 ⌣ f .(x) I .P V .A . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 42 / 47
    • Step 3: Synthesis . − .. . 0 . + ∞ − .. . .′ f ↘ . 2 . − ↗ . 0 ↘ . . m . onotonicity − . − .. 0 . + + ∞ + .. . + .′′ f . ⌢ . 3 − . ⌣ . . 0 ⌣ c . oncavity 0 . − . 2/9 − . 1/4 0 .. ∞ .. 0f .. . . − . ∞ . . 3 − − − . 2 − + . 1 . 0 . . + ∞s . . hape of f . . . . . .
    • Step 3: Synthesis . − .. . 0 . + ∞ − .. . .′ f ↘ . 2 . − ↗ . 0 ↘ . . m . onotonicity − . − .. 0 . + + ∞ + .. . + .′′ f . ⌢ . 3 − . ⌣ . . 0 ⌣ c . oncavity 0 . − . 2/9 − . 1/4 0 .. ∞ .. 0f .. . . − . ∞ . . 3 − − − . 2 − + . 1 . 0 . . + ∞s . . hape of f H . A . . . . . .
    • Step 3: Synthesis . − .. . 0 . + ∞ − .. . .′ f ↘ . 2 . − ↗ . 0 ↘ . . m . onotonicity − . − .. 0 . + + ∞ + .. . + .′′ f . ⌢ . 3 − . ⌣ . . 0 ⌣ c . oncavity 0 . − . 2/9 − . 1/4 0 .. ∞ .. 0f .. . . − . ∞ . . 3 − − − . 2 − + . 1 . 0 . . + ∞s . . hape of f . A . H . . . . . .
    • Step 3: Synthesis . − .. . 0 . + ∞ − .. . .′ f ↘ . 2 . − ↗ . 0 ↘ . . m . onotonicity − . − .. 0 . + + ∞ + .. . + .′′ f . ⌢ . 3 − . ⌣ . . 0 ⌣ c . oncavity 0 . − . 2/9 − . 1/4 0 .. ∞ .. 0f .. . . − . ∞ . . 3 − − − . 2 − + . 1 . 0 . . + ∞s . . hape of f . A . .P H I . . . . . .
    • Step 3: Synthesis . − .. . 0 . + ∞ − .. . .′ f ↘ . 2 . − ↗ . 0 ↘ . . m . onotonicity − . − .. 0 . + + ∞ + .. . + .′′ f . ⌢ . 3 − . ⌣ . . 0 ⌣ c . oncavity 0 . − . 2/9 − . 1/4 0 .. ∞ .. 0f .. . . − . ∞ . . 3 − − − . 2 − + . 1 . 0 . . + ∞s . . hape of f . A . .P . H I . . . . . .
    • Step 3: Synthesis . − .. . 0 . + ∞ − .. . .′ f ↘ . 2 . − ↗ . 0 ↘ . . m . onotonicity − . − .. 0 . + + ∞ + .. . + .′′ f . ⌢ . 3 − . ⌣ . . 0 ⌣ c . oncavity 0 . − . 2/9 − . 1/4 0 .. ∞ .. 0f .. . . − . ∞ . . 3 − − − . 2 − + . 1 . 0 . . + ∞s . . hape of f . A . .P . . in H I m . . . . . .
    • Step 3: Synthesis . − .. . 0 . + ∞ − .. . .′ f ↘ . 2 . − ↗ . 0 ↘ . . m . onotonicity − . − .. 0 . + + ∞ + .. . + .′′ f . ⌢ . 3 − . ⌣ . . 0 ⌣ c . oncavity 0 . − . 2/9 − . 1/4 0 .. ∞ .. 0f .. . . − . ∞ . . 3 − − − . 2 − + . 1 . 0 . . + ∞s . . hape of f . A . .P . . in . H I m . . . . . .
    • Step 3: Synthesis . − .. . 0 . + ∞ − .. . .′ f ↘ . 2 . − ↗ . 0 ↘ . . m . onotonicity − . − .. 0 . + + ∞ + .. . + .′′ f . ⌢ . 3 − . ⌣ . . 0 ⌣ c . oncavity 0 . − . 2/9 − . 1/4 0 .. ∞ .. 0f .. . . − . ∞ . . 3 − − − . 2 − + . 1 . 0 . . + ∞s . . hape of f . A . .P . . in . H I m 0 . . . . . . .
    • Step 3: Synthesis . − .. . 0 . + ∞ − .. . .′ f ↘ . 2 . − ↗ . 0 ↘ . . m . onotonicity − . − .. 0 . + + ∞ + .. . + .′′ f . ⌢ . 3 − . ⌣ . . 0 ⌣ c . oncavity 0 . − . 2/9 − . 1/4 0 .. ∞ .. 0f .. . . − . ∞ . . 3 − − − . 2 − + . 1 . 0 . . + ∞s . . hape of f . A . .P . . in . H I m 0 . . . . . . . .
    • Step 3: Synthesis . − .. . 0 . + ∞ − .. . .′ f ↘ . 2 . − ↗ . 0 ↘ . . m . onotonicity − . − .. 0 . + + ∞ + .. . + .′′ f . ⌢ . 3 − . ⌣ . . 0 ⌣ c . oncavity 0 . − . 2/9 − . 1/4 0 .. ∞ .. 0f .. . . − . ∞ . . 3 − − − . 2 − . 1 . + .0 . + ∞s . . hape of f . A . .P . . in . H I m 0 . .A . V . . . . . .
    • Step 3: Synthesis . − .. . 0 . + ∞ − .. . .′ f ↘ . 2 . − ↗ . 0 ↘ . . m . onotonicity − . − .. 0 . + + ∞ + .. . + .′′ f . ⌢ . 3 − . ⌣ . . 0 ⌣ c . oncavity 0 . − . 2/9 − . 1/4 0 .. ∞ .. 0f .. . . − . ∞ . . 3 − − − . 2 − . 1 . + .0 . + ∞s . . hape of f . A . .P . . in . H I m 0 . .A . . V . . . . . .
    • Step 3: Synthesis . − .. . 0 . + ∞ − .. . .′ f ↘ . 2 . − ↗ . 0 ↘ . . m . onotonicity − . − .. 0 . + + ∞ + .. . + .′′ f . ⌢ . 3 − . ⌣ . . 0 ⌣ c . oncavity 0 . − . 2/9 − . 1/4 0 .. ∞ .. 0f .. . . − . ∞ . . 3 − − − . 2 − . 1 . + .0 . ∞s + . . hape of f . A . .P . . in . H I m 0 . .A . . A . V H . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 43 / 47
    • Step 4: Graph y . . x . . . . −3, −2/9) . −2, −1/4) ( ( . − − 0 . 2/9 . 1/4 .. 0 ∞ .. 0f .. . . − −− . ∞. . 3 . 2 . 1 . . . − − + 0 + ∞s . . hape of f . A . .P . . in . . . . A . H I m 0 V H . A . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 44 / 47
    • Problem Graph f(x) = cos x − x . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 45 / 47
    • Problem Graph f(x) = cos x − x y . . x . . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 45 / 47
    • Problem Graph f(x) = x ln x2 . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 46 / 47
    • Problem Graph f(x) = x ln x2 y . . x . . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 46 / 47
    • Graphing Checklist To graph a function f, follow this plan: 0. Find when f is positive, negative, zero, not defined. 1. Find f′ and form its sign chart. Conclude information about increasing/decreasing and local max/min. 2. Find f′′ and form its sign chart. Conclude concave up/concave down and inflection. 3. Put together a big chart to assemble monotonicity and concavity data 4. Graph! . . . . . . V63.0121, Calculus I (NYU) Section 4.4 Curve Sketching April 1, 2010 47 / 47