
Concept explainers
(a)
To find:The intervals of increase or decrease.
(a)

Explanation of Solution
Given: B(x)=3x2/3−x
Calculate the first derivative of the given function.
B′(x)=3⋅23x−1/3−10=2−x1/3x1/3x=0, 8
The function is decreasing in the interval (−∞, 0) .
The function is increasing in the interval (0, 8)
The function is decreasing in the interval (8, ∞) .
(b)
To find :The
(b)

Explanation of Solution
The function is decreasing in the interval (−∞, 0) .
The function is increasing in the interval (0, 8)
The function is decreasing in the interval (8, ∞)
The value of B′ is changing from negative to positive at x=0
So, the local
The value of B′ is changing from positive to negative at x=8
So, the local maxima is (8, 4)
(c)
To find :The concavity of the function and the point of inflection.
(c)

Explanation of Solution
Calculate the second derivative of the given function.
B″(x)=−23x−4/3
B″(x)<0 for the interval −∞<x<0 . So, the graph is concave down for (−∞, 0)
B″(x)<0 for the interval 0<x<∞ . So, the graph is concave down for (0, ∞)
There is no change of sign so no inflection point.
(d)
To sketch :The graph of result obtained from part (a) to (c).
(d)

Explanation of Solution
Given: B(x)=3x2/3−x
The graph of the result obtained is shown in figure below.
Figure (1)
Therefore, the required graph is shown in figure (1).
Chapter 4 Solutions
Single Variable Calculus: Concepts and Contexts, Enhanced Edition
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