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

Explanation of Solution
Given: h(x)=(x+1)5−5x−2
Calculate the first derivative of the given function.
h′(x)=5(x+1)4−50=5(x+1)4−51=(x+1)4x=0, 2
The function is increasing in the interval (−∞, −2)
The function is decreasing in the interval (−2, o) .
The function is increasing in the interval (0, ∞)
(b)
To find :The
(b)

Explanation of Solution
The function is increasing in the interval (−∞, −2)
The function is decreasing in the interval (−2, o) .
The function is increasing in the interval (0, ∞)
The value of f′ is changing from positive to negative at x=−2 .
So, the local
The value of f′ is changing from negative to positive at x=0 .
So, the local minima is (0, f(0))=(0, −1)
(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.
h″(x)=20(x+1)30=20(x+1)5x=−1
f″(x)<0 for the interval x<−1 . So, the graph is concave down for (−∞, −1) .
f″(x)>0 for the interval x>−1 . So, the graph is concave upfor (−1, ∞) .
The inflection point is (−1, 3) .
(d)
To sketch :The graph of result obtained from part (a) to (c).
(d)

Explanation of Solution
Given: h(x)=(x+1)5−5x−2
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
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