MATH 12002 - CALCULUS I§3.4: Curve Sketching - Another Example

Professor Donald L. White

Department of Mathematical SciencesKent State University

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Example 2

Example 2

Let

f (x) =2x2

(x + 1)2.

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 / 13

Example 2

We need to determine the signs of f ′ and f ′′ for

f (x) =2x2

(x + 1)2= 2

[x

x + 1

]2

.

First,

f ′(x) = 4

[x

x + 1

]1

· 1 · (x + 1)− x · 1(x + 1)2

= 4

[x

x + 1

]· 1

(x + 1)2

=4x

(x + 1)3.

Hence f ′(x) is 0 when x = 0 and f ′(x) is undefined when x = −1.

D.L. White (Kent State University) 3 / 13

Example 2

Using f ′(x) =4x

(x + 1)3, we have

f ′′(x) =4(x + 1)3 − 4x · 3(x + 1)2

(x + 1)6

=4(x + 1)2[(x + 1)− x · 3]

(x + 1)6

=4(1− 2x)

(x + 1)4

=8(1

2 − x)

(x + 1)4.

Hence f ′′(x) = 0 when x = 12 and f ′′(x) is undefined when x = −1.

D.L. White (Kent State University) 4 / 13

Example 2

f (x) = 2x2

(x+1)2, f ′(x) = 4x

(x+1)3, f ′′(x)

8( 12−x)

(x+1)4

−1 012

4x

(x + 1)3

f ′(x)

8( 12 − x)

(x + 1)4

f ′′(x)

Inc-Dec

Concave

Shape

0− − + +

0− + + +

X 0+ − + +

0+ + + −

0+ + + +

X 0+ + + −-

-

XI D I I

�

�

XU U U D � �MIN

INF

D.L. White (Kent State University) 5 / 13

Example 2

f (x) =2x2

(x + 1)2, f ′(x) =

4x

(x + 1)3, f ′′(x) =

8(12 − x)

(x + 1)4

−1 0 12

Inc-Dec

Concave

Shape

XI D I IMIN

XU U U DINF �f is increasing on (−∞,−1) ∪ (0,∞);f is decreasing on (−1, 0);f has a local minimum at x = 0.f is concave up on (−∞,−1) ∪ (−1, 1

2);f is concave down on (1

2 ,∞);f has an inflection point at x = 1

2 .

D.L. White (Kent State University) 6 / 13

Example 2

In order to sketch the graph of f , we will need to plot the points whose xcoordinates are in the sign chart:

f (x) =2x2

(x + 1)2is undefined at x = −1

f (0) =2(02)

(0 + 1)2=

0

1= 0

f (12) =

2(12)2

(12 + 1)2

=1/2

9/4=

1

2· 4

9=

2

9

Hence the points (0, 0) and (12 , 2

9) are on the graph.

D.L. White (Kent State University) 7 / 13

Example 2

Finally, we find the horizontal and vertical asymptotes.

Horizontal Asymptotes: f (x) =2x2

(x + 1)2=

2x2

x2 + 2x + 1and the numerator and denominator of f have the same degree.Hence the horizontal asymptote is y = 2, the quotient of the leadingcoefficients.

Vertical Asymptotes: f (x) =2x2

(x + 1)2

and the denominator is 0, while the numerator is not, when x = −1.Hence x = −1 is the vertical asymptote.

D.L. White (Kent State University) 8 / 13

Example 2

−1 0 12

Inc-Dec

Concave

Shape

XI D I IMIN

XU U U DINF �

-�

6

?−1

1

2

3

4

5

6

−3 −2 −1 1 2 3q

(0, 0), MIN���

q( 1

2, 2

9), INF

@@I

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Example 2

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Example 2

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Example 2

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Example 2

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