Single Variable Calculus: Concepts and Contexts, Enhanced Edition

4th Edition

ISBN: 9781337687805

Author: James Stewart

Publisher: Cengage Learning

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Question

Chapter 4.3, Problem 34E

(a)

To determine

To find : The vertical and horizontal asymptotes.

(a)

Expert Solution

Explanation of Solution

Given: The function is $f(x)=x2(x−2)2$

Consider the function.

$f(x)=x2(x−2)2$

The horizontal asymptote is calculated as,

$limx→∞f(x)=limx→∞x2(x−2)2=limx→∞x2x2(1−2x)2=limx→∞11−2x=11−0=1$

So, the horizontal asymptote is $f(x)=1$ .

The vertical asymptote is calculated as,

$limx→2+f(x)=limx→2+x2(x−2)2=limx→2+x2limx→2+(x−2)2$

But,

$limx→2+x2=4$

And,

$limx→2+(x−2)2=0+$

Thus,

$limx→2+f(x)=∞$

So, the vertical asymptotes is $x=2$ .

(b)

To determine

To find : The interval of increase or decrease.

(b)

Expert Solution

Explanation of Solution

Given: The function is $f(x)=x2(x−2)2$

Consider the function.

$f(x)=x2(x−2)2$

Differentiate the above expression with respect to $x$ .

$f′(x)=(x−2)2(2x)−(x2)2(x−2)(1)((x−2)2)2=(2x2−4x)−(2x2)(x−2)3=−4x(x−2)3$

The function is decreasing in the interval $(−∞, 0)$ .

The function is increasing in the interval $(0, 2)$

The function is decreasing in the interval $(2, ∞)$

(c)

To determine

To find : The local maximum and minimum values..

(c)

Expert Solution

Explanation of Solution

Given: The function is $f(x)=x2(x−2)2$

The function is decreasing in the interval $(−∞, 0)$ .

The function is increasing in the interval $(0, 2)$

The function is decreasing in the interval $(2, ∞)$

So, the function is minimum at $x=0$ and the local minima is $(0, f(0))=(0,0)$ .

There is no local maxima.

(d)

To determine

To find : The intervals of concavity and the inflection points.

(d)

Expert Solution

Explanation of Solution

Given: The function is $f(x)=x2(x−2)2$

Differentiate $f′(x)$ with respect to $x$ .

$f″(x)=(x−2)(−4)−(−4x)(3)(x−2)4=−4x+8+12x(x−2)4=8(1+x)(x−2)4$

The value of $f″(x)$ is .negative in for $x<−1$ So the concavity is downward in the interval $(−∞, −1)$ .

The value of $f″(c)$ is positive for $−1<x<2$ . So the concavity is upwars in the interval $(−1, 1)$ .

The function has inflection point $(−1, 19)$

(e)

To determine

To sketch : The graph of the function.

(e)

Expert Solution

Explanation of Solution

Given: The function is $f(x)=x2(x−2)2$

Consider the function.

$f(x)=x2(x−2)2$

The graph of the above function is shown in figure below.

Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 4.3, Problem 34E

Figure (1)

Therefore, the graph of the function is shown in Figure (1).

Chapter 4.3, Problem 33E

Chapter 4.3, Problem 35E

Chapter 4 Solutions

Single Variable Calculus: Concepts and Contexts, Enhanced Edition

Ch. 4.1 - Prob. 1E Ch. 4.1 - Prob. 2E Ch. 4.1 - Prob. 3E Ch. 4.1 - Prob. 4E Ch. 4.1 - Prob. 5E Ch. 4.1 - Prob. 6E Ch. 4.1 - Prob. 7E Ch. 4.1 - Prob. 8E Ch. 4.1 - Prob. 9E Ch. 4.1 - Prob. 10E

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