MATH 12002 - CALCULUS I 3.4: Curve Sketching Professor Donald L. - - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I §3.4: Curve Sketching

Professor Donald L. White

Department of Mathematical Sciences Kent State University

D.L. White (Kent State University) 1 / 12

Example 1

Example 1

Let f (x) = x x2 − 9. Determine intervals where f is increasing, intervals where f is decreasing, the location of all local maxima and minima, intervals where f is concave up, intervals where f is concave down, the location of all inflection points, and all vertical and horizontal asymptotes.

D.L. White (Kent State University) 2 / 12

Example 1

We need to determine the signs of f ′ and f ′′ for f (x) =

x x2−9.

First, f ′(x) = 1 · (x2 − 9) − x · 2x (x2 − 9)2 = x2 − 9 − 2x2 (x2 − 9)2 = −x2 − 9 (x2 − 9)2 = −(x2 + 9) (x + 3)2(x − 3)2 Hence f ′(x) is never 0 and f ′(x) is undefined when x = −3 or x = 3.

D.L. White (Kent State University) 3 / 12

Example 1

Using f ′(x) = −(x2+9)

(x2−9)2 , we have

f ′′(x) = −2x(x2 − 9)2 + (x2 + 9) · 2(x2 − 9) · 2x (x2 − 9)4 = (x2 − 9)(2x)

• −(x2 − 9) + (x2 + 9) · 2
• (x2 − 9)4

= 2x ·

• −x2 + 9 + 2x2 + 18
• (x2 − 9)3

= 2x · (x2 + 27) (x + 3)3(x − 3)3 . Hence f ′′(x) = 0 when x = 0 and f ′′(x) is undefined when x = −3 or x = 3.

D.L. White (Kent State University) 4 / 12

Example 1

f (x) =

x x2−9, f ′(x) = −(x2+9) (x+3)2(x−3)2 , f ′′(x) = 2x·(x2+27) (x+3)3(x−3)3

−3 3 −1 x2 + 9 (x − 3)2 (x + 3)2 f ′(x) 2x x2 + 27 (x − 3)3 (x + 3)3 f ′′(x)

Inc-Dec Concave Shape

− − − − + + + + + + + + + + + +

X X

− − − − − − + + + + + + − − − + − + + +

X X

− + − + ✲ ✲

X X

D D D D ✛ ✛

X X

D U D U ✟ ✡ ✟ ✡

INF D.L. White (Kent State University) 5 / 12

Example 1

f (x) = x x2 − 9, f ′(x) = −(x2 + 9) (x + 3)2(x − 3)2 , f ′′(x) = 2x · (x2 + 27) (x + 3)3(x − 3)3

−3 3

Inc-Dec Concave Shape

X X

D D D D

X X

D U D U

INF

✟ ✡ ✟ ✡

f is decreasing on (−∞, −3) ∪ (−3, 3) ∪ (3, ∞); f has no local minimum or local maximum. f is concave up on (−3, 0) ∪ (3, ∞); f is concave down on (−∞, −3) ∪ (0, 3); f has an inflection point at x = 0.

D.L. White (Kent State University) 6 / 12

Example 1

In order to sketch the graph of f , we will need to plot the points whose x coordinates are in the sign chart: f (x) =

x x2−9 is undefined at x = −3 and x = 3;

f (0) =

02−9 = −9 = 0.

Hence the point (0, 0) is on the graph. Horizontal Asymptotes: f (x) =

x x2−9 and the numerator of f has lower

degree than the denominator. Hence y = 0 is the horizontal asymptote. Vertical Asymptotes: f (x) =

x (x+3)(x−3) and the denominator is 0 when

x = −3 and when x = 3, and the numerator is not. Hence x = −3 and x = 3 are vertical asymptotes.

D.L. White (Kent State University) 7 / 12

Example 1

−3 3

Inc-Dec Concave Shape

X X

D D D D

X X

D U D U

INF

✟ ✡ ✟ ✡ ✲ ✛ ✻ ❄

−6 −3 3 6

q

(0, 0), INF D.L. White (Kent State University) 8 / 12

Example 1

D.L. White (Kent State University) 9 / 12

Example 1

D.L. White (Kent State University) 10 / 12

Example 1

D.L. White (Kent State University) 11 / 12

Example 1

D.L. White (Kent State University) 12 / 12