(a) Explanation Given: f ( x ) = ln ( x 4 + 27 ) Calculate the first derivative of the given function. f ′ ( x ) = 4 x 3 x 4 + 27 Equate to zero. 0 = 4 x 3 x 4 + 27 x = 0 The function is decreasing in the interval ( − ∞ , 0 ) . The function is increasing in the interval ( 0 , ∞ ) (b) To determine To find : The local maxima and local minima . Explanation The function is decreasing in the interval ( − ∞ , 0 ) . The function is increasing in the interval ( 0 , ∞ ) The value of f ′ is changing from negative to positive at x = 0 So, the local minima is ( 0 , ln 27 ) There is no local maxima . (c) To determine To find : The concavity of the function and the point of inflection. Explanation Calculate the second derivative of the given function. f ″ ( x ) = 324 x 2 − 4 x 6 ( x 4 + 27 ) 2 = 4 x 2 ( x 4 − 81 ) ( x 4 + 27 ) 2 4 x 2 = 0 x = 0 f ″ ( x ) < 0 for the interval − ∞ < x < − 3 . So, the graph is concave down for ( − ∞ , − 3 ) f ″ ( x ) > 0 for the interval − 3 < x < 0 . So, the graph is concave up for ( − 3 , 0 ) f ″ ( x ) > 0 for the interval 0 < x < 3 . So, the graph is concave up for ( 0 , 3 ) f ″ ( x ) < 0 for the interval 3 < x < ∞ . So, the graph is concave down for ( 3 , ∞ ) The inflection points are ( − 3. 4.68 ) , ( 0 , 3.30 ) and ( 3 , 4.68 ) (d) To determine To sketch : The graph of result obtained from part (a) to (c). Explanation Given: f ( x ) = ln ( x 4 + 27 ) The graph of the result obtained is shown in figure below. Figure (1) Therefore, the required graph is shown in figure (1).

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 30E

(a)

To determine

To find:The intervals of increase or decrease.

(a)

Expert Solution

Explanation of Solution

Given: $f(x)=ln(x4+27)$

Calculate the first derivative of the given function.

$f′(x)=4x3x4+27$

Equate to zero.

$0=4x3x4+27x=0$

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

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

(b)

To determine

To find :The local maxima and local minima.

(b)

Expert Solution

Explanation of Solution

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

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

The value of $f′$ is changing from negative to positive at $x=0$

So, the local minima is $(0, ln27)$

There is no local maxima.

(c)

To determine

To find :The concavity of the function and the point of inflection.

(c)

Expert Solution

Explanation of Solution

Calculate the second derivative of the given function.

$f″(x)=324x2−4x6(x4+27)2=4x2(x4−81)(x4+27)24x2=0x=0$

$f″(x)<0$ for the interval $−∞<x<−3$ . So, the graph is concave down for $(−∞, −3)$

$f″(x)>0$ for the interval $−3<x<0$ . So, the graph is concave up for $(−3, 0)$

$f″(x)>0$ for the interval $0<x<3$ . So, the graph is concave up for $(0, 3)$

$f″(x)<0$ for the interval $3<x<∞$ . So, the graph is concave down for $(3, ∞)$

The inflection points are $(−3. 4.68), (0, 3.30) and (3, 4.68)$

(d)

To determine

To sketch :The graph of result obtained from part (a) to (c).

(d)

Expert Solution

Explanation of Solution

Given: $f(x)=ln(x4+27)$

The graph of the result obtained is shown in figure below.

Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 4.3, Problem 30E

Figure (1)

Therefore, the required graph is shown in figure (1).

Chapter 4.3, Problem 29E

Chapter 4.3, Problem 31E

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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