Lesson 21: Curve Sketching (handout)
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The document contains lecture notes on curve sketching from a Calculus I class. It discusses using the first and second derivative tests to determine properties of a function like monotonicity, concavity, maxima, minima, and points of inflection in order to sketch the graph of the function. It then provides an example of using these tests to sketch the graph of the cubic function f(x) = 2x^3 - 3x^2 - 12x.
Lesson 21: Curve Sketching (handout)
- 1. . V63.0121.001: Calculus I . Sec on 4.4: Curve Sketching . April 13, 2011 Notes Sec on 4.4 Curve Sketching V63.0121.001: Calculus I Professor Ma hew Leingang New York University April 13, 2011 . . Notes Announcements Quiz 4 on Sec ons 3.3, 3.4, 3.5, and 3.7 this week (April 14/15) Quiz 5 on Sec ons 4.1–4.4 April 28/29 Final Exam Thursday May 12, 2:00–3:50pm I am teaching Calc II MW 2:00pm and Calc III TR 2:00pm both Fall ’11 and Spring ’12 . . Notes Objectives given a func on, graph it completely, indica ng zeroes (if easy) asymptotes if applicable cri cal points local/global max/min inflec on points . . . 1 .
- 2. . V63.0121.001: Calculus I . Sec on 4.4: Curve Sketching . April 13, 2011 Notes Why? Graphing func ons is like dissec on … or diagramming sentences You can really know a lot about a func on when you know all of its anatomy. . . Notes The Increasing/Decreasing Test Theorem (The Increasing/Decreasing Test) If f′ > 0 on (a, b), then f is increasing on (a, b). If f′ < 0 on (a, b), then f is decreasing on (a, b). Example f(x) f′ (x) f(x) = x3 + x2 f′ (x) = 3x2 + 2x . . . Notes Testing for Concavity Theorem (Concavity Test) If f′′ (x) > 0 for all x in (a, b), then the graph of f is concave upward on (a, b) If f′′ (x) < 0 for all x in (a, b), then the graph of f is concave downward on (a, b). Example f′′ (x) f′ (x) f(x) f(x) = x3 + x2 f′ (x) = 3x2 + 2x . f′′ (x) = 6x + 2 . . . 2 .
- 3. . V63.0121.001: Calculus I . Sec on 4.4: Curve Sketching . April 13, 2011 Notes Graphing Checklist To graph a func on f, follow this plan: 0. Find when f is posi ve, nega ve, zero, not defined. 1. Find f′ and form its sign chart. Conclude informa on about increasing/decreasing and local max/min. 2. Find f′′ and form its sign chart. Conclude concave up/concave down and inflec on. . . Notes Graphing Checklist To graph a func on f, follow this plan: 3. Put together a big chart to assemble monotonicity and concavity data 4. Graph! . . Notes Outline Simple examples A cubic func on A quar c func on More Examples Points of nondifferen ability Horizontal asymptotes Ver cal asymptotes Trigonometric and polynomial together Logarithmic . . . 3 .
- 4. . V63.0121.001: Calculus I . Sec on 4.4: Curve Sketching . April 13, 2011 Notes Graphing a cubic Example Graph f(x) = 2x3 − 3x2 − 12x. (Step 0) First, let’s find the zeros. We can at least factor out one power of x: f(x) = x(2x2 − 3x − 12) so f(0) = 0. The other factor is a quadra c, so we the other two roots are √ √ 3 ± 32 − 4(2)(−12) 3 ± 105 x= = 4 4 . It’s OK to skip this step for now since the roots are so complicated. . Notes Step 1: Monotonicity f(x) = 2x3 − 3x2 − 12x =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: − .− + x−2 2 − + + x+1 −1 f′ (x) + − + ↗−1 ↘ 2 ↗ f(x) max min . . Notes Step 2: Concavity f′ (x) = 6x2 − 6x − 12 =⇒ f′′ (x) = 12x − 6 = 6(2x − 1) Another sign chart: . −− ++ f′′ (x) ⌢ 1/2 ⌣ f(x) IP . . . 4 .
- 5. . V63.0121.001: Calculus I . Sec on 4.4: Curve Sketching . April 13, 2011 Notes Step 3: One sign chart to rule them all Remember, f(x) = 2x3 − 3x2 − 12x. + −. − + f′ (x) ↗−1 ↘ ↘ 2 ↗ monotonicity −− −− ++ ++ f′′ (x) ⌢ ⌢ 1/2 ⌣ ⌣ concavity 7 −61/2 −20 f(x) −1 1/2 2 shape of f max IP min . . Notes monotonicity and concavity increasing, decreasing, concave concave down down II I . III IV decreasing, increasing, concave concave up up . . Notes Step 3: One sign chart to rule them all Remember, f(x) = 2x3 − 3x2 − 12x. + −. − + f′ (x) ↗−1 ↘ ↘ 2 ↗ monotonicity −− −− ++ ++ f′′ (x) ⌢ ⌢ 1/2 ⌣ ⌣ concavity 7 −61/2 −20 f(x) −1 1/2 2 shape of f max IP min . . . 5 .
- 6. . V63.0121.001: Calculus I . Sec on 4.4: Curve Sketching . April 13, 2011 f(x) Notes Step 4: Graph f(x) = 2x3 − 3x2 − 12x ( √ ) (−1, 7) 3− 105 4 ,0 (0, 0) . ( x√ ) (1/2, −61/2) 3+ 105 4 ,0 (2, −20) 7 −61/2 −20 f(x) −1 1/2 2 shape of f max IP min . . Notes Graphing a quartic Example Graph f(x) = x4 − 4x3 + 10 (Step 0) We know f(0) = 10 and lim f(x) = +∞. Not too many x→±∞ other points on the graph are evident. . . Notes Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. +0 . + + 4x2 0 − − 0+ (x − 3) 3 ′ −0 − 0 + f (x) ↘0 ↘ 3 ↗ f(x) min . . . 6 .
- 7. . V63.0121.001: Calculus I . Sec on 4.4: Curve Sketching . April 13, 2011 Notes Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: −0. + + 12x 0 − 0 +− x−2 2 f′′ (x) ++0 −− 0 ++ ⌣0 ⌢ 2 ⌣ f(x) IP IP . . Notes Step 3: Grand Unified Sign Chart . Remember, f(x) = x4 − 4x3 + 10. −0 − −0+ f′ (x) ↘0 ↘ ↘3↗ monotonicity f′′ (x) ++0 −− 0++ ++ ⌣0 ⌢ 2⌣ ⌣ concavity 10 −6 −17 f(x) 0 2 3 shape IP IP min . . y Notes Step 4: Graph f(x) = x4 − 4x3 + 10 (0, 10) . x (2, −6) (3, −17) 10 −6 −17 f(x) 0 2 3 shape IP IP min . . . 7 .
- 8. . V63.0121.001: Calculus I . Sec on 4.4: Curve Sketching . April 13, 2011 Notes Outline Simple examples A cubic func on A quar c func on More Examples Points of nondifferen ability Horizontal asymptotes Ver cal asymptotes Trigonometric and polynomial together Logarithmic . . Notes Graphing a function with a cusp Example √ Graph f(x) = x + |x| This func on looks strange because of the absolute value. But whenever we become nervous, we can just take cases. . . Notes Step 0: Finding Zeroes √ f(x) = x + |x| First, look at f by itself. We can tell that f(0) = 0 and that f(x) > 0 if x is posi ve. Are there nega ve numbers which are zeroes for f? √ √ x + −x = 0 =⇒ −x = −x −x = x2 =⇒ x2 + x = 0 The only solu ons are x = 0 and x = −1. . . . 8 .
- 9. . V63.0121.001: Calculus I . Sec on 4.4: Curve Sketching . April 13, 2011 Notes Step 0: Asymptotic behavior √ f(x) = x + |x| lim f(x) = ∞, because both terms tend to ∞. x→∞ lim f(x) is indeterminate of the form −∞ + ∞. It’s the same x→−∞ √ as lim (−y + y) y→+∞ √ √ √ y+y lim (−y + y) = lim ( y − y) · √ y→+∞ y→∞ y+y y − y2 = lim √ = −∞ y→∞ y+y . . Notes Step 1: The derivative √ Remember, f(x) = x + |x|. To find f′ , first assume x > 0. Then d ( √ ) 1 f′ (x) = x+ x =1+ √ dx 2 x No ce f′ (x) > 0 when x > 0 (so no cri cal points here) lim+ f′ (x) = ∞ (so 0 is a cri cal point) x→0 lim f′ (x) = 1 (so the graph is asympto c to a line of slope 1) x→∞ . . Notes Step 1: The derivative √ Remember, f(x) = x + |x|. If x is nega ve, we have d ( √ ) 1 f′ (x) = x + −x = 1 − √ dx 2 −x No ce lim− f′ (x) = −∞ (other side of the cri cal point) x→0 lim f′ (x) = 1 (asympto c to a line of slope 1) x→−∞ ′ f (x) = 0 when 1 √ 1 1 1 1− √ = 0 =⇒ −x = =⇒ −x = =⇒ x = − 2 −x 2 4 4 . . . 9 .
- 10. . V63.0121.001: Calculus I . Sec on 4.4: Curve Sketching . April 13, 2011 Notes Step 1: Monotonicity 1 1 + √ if x > 0 ′ f (x) = 2 x 1 − √ 1 if x < 0 2 −x We can’t make a mul -factor sign chart because of the absolute value, but we can test points in between cri cal points. + 0− ∞ + f′ (x) . ↗ − 1↘ 0 4 ↗ f(x) max min . . Notes Step 2: Concavity ( ) d 1 1 If x > 0, then f′′ (x) = 1 + x−1/2 = − x−3/2 This is dx 2 4 nega ve whenever x > 0. ( ) d 1 1 If x < 0, then f′′ (x) = 1 − (−x)−1/2 = − (−x)−3/2 dx 2 4 which is also always nega ve for nega ve x. 1 In other words, f′′ (x) = − |x|−3/2 . 4 Here is the sign chart: −− −∞ −− f′′ (x) . ⌢ 0 ⌢ f(x) . . Notes Step 3: Synthesis Now we can put these things together. √ f(x) = x + |x| +1 + 0− ∞ + f′ +1 (x) . ↗ ↗ − 1↘ 0 ↗ ↗monotonicity f′′ −∞ −− −− 4 −∞ −− −∞ (x) ⌢ ⌢ 1 ⌢0 ⌢ ⌢concavity −∞ 0 4 0 +∞f(x) −1 −4 0 1 shape zero max min . . . 10 .
- 11. . V63.0121.001: Calculus I . Sec on 4.4: Curve Sketching . April 13, 2011 Notes Graph √ f(x) = x + |x| (− 1 , 1 ) 4 4 (−1, 0) . x (0, 0) 1 −∞ 0 4 0 +∞ x −1 −1 0 shape 4 zero max min . . Example with Horizontal Notes Asymptotes Example Graph f(x) = xe−x 2 . . Notes Step 1: Monotonicity . . . 11 .
- 12. . V63.0121.001: Calculus I . Sec on 4.4: Curve Sketching . April 13, 2011 Notes Step 2: Concavity . . Notes Step 3: Synthesis . . Notes Step 4: Graph f(x) x . . . . 12 .
- 13. . V63.0121.001: Calculus I . Sec on 4.4: Curve Sketching . April 13, 2011 Notes Example with Vertical Asymptotes Example 1 1 Graph f(x) = + x x2 . . Notes Step 0 . . Notes Step 1: Monotonicity . . . 13 .
- 14. . V63.0121.001: Calculus I . Sec on 4.4: Curve Sketching . April 13, 2011 Notes Step 2: Concavity . . Notes Step 3: Synthesis . . Notes Step 4: Graph y . x . . . 14 .
- 15. . V63.0121.001: Calculus I . Sec on 4.4: Curve Sketching . April 13, 2011 Trigonometric and polynomial Notes together Problem Graph f(x) = cos x − x . . Notes Step 0: intercepts and asymptotes . . Notes Step 1: Monotonicity . . . 15 .
- 16. . V63.0121.001: Calculus I . Sec on 4.4: Curve Sketching . April 13, 2011 Notes Step 2: Concavity . . Notes Step 3: Synthesis . . Notes Step 4: Graph f(x) = cos x − x y . x . . . 16 .
- 17. . V63.0121.001: Calculus I . Sec on 4.4: Curve Sketching . April 13, 2011 Notes Logarithmic Problem Graph f(x) = x ln x2 . . Step 0: Intercepts and Notes Asymptotes . . Notes Step 1: Monotonicity . . . 17 .
- 18. . V63.0121.001: Calculus I . Sec on 4.4: Curve Sketching . April 13, 2011 Notes Step 2: Concavity . . Notes Step 3: Synthesis . . Notes Step 4: Graph y . x . . . 18 .
- 19. . V63.0121.001: Calculus I . Sec on 4.4: Curve Sketching . April 13, 2011 Notes Summary Graphing is a procedure that gets easier with prac ce. Remember to follow the checklist. . . Notes . . Notes . . . 19 .