Chapter 4.3, Problem 25E

(a)

To determine

To find:The intervals of increase or decrease.

Explanation of Solution

Given: $h\left(x\right)={\left(x+1\right)}^5-5x-2$

Calculate the first derivative of the given function.

  $\begin{array}{c}{h}^{\prime }\left(x\right)=5{\left(x+1\right)}^4-5\\ 0=5{\left(x+1\right)}^4-5\\ 1={\left(x+1\right)}^4\\ x=0,2\end{array}$

The function is increasing in the interval $\left(-\infty ,-2\right)$

The function is decreasing in the interval $\left(-2,o\right)$ .

The function is increasing in the interval $\left(0,\infty \right)$

(b)

To determine

To find :The local maxima and local minima.

Explanation of Solution

The function is increasing in the interval $\left(-\infty ,-2\right)$

The function is decreasing in the interval $\left(-2,o\right)$ .

The function is increasing in the interval $\left(0,\infty \right)$

The value of $f^{\prime }$ is changing from positive to negative at $x=-2$ .

So, the local minima is $\left(-2,f\left(-2\right)\right)=\left(-2,7\right)$

The value of $f^{\prime }$ is changing from negative to positive at $x=0$ .

So, the local minima is $\left(0,f\left(0\right)\right)=\left(0,-1\right)$

(c)

To determine

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

Explanation of Solution

Calculate the second derivative of the given function.

  $\begin{array}{c}{h}^{″}\left(x\right)=20{\left(x+1\right)}^3\\ 0=20{\left(x+1\right)}^5\\ x=-1\end{array}$

  $f^{″}\left(x\right)<0$ for the interval $x<-1$ . So, the graph is concave down for $\left(-\infty ,-1\right)$ .

  $f^{″}\left(x\right)>0$ for the interval $x>-1$ . So, the graph is concave upfor $\left(-1,\infty \right)$ .

The inflection point is $\left(-1,3\right)$ .

(d)

To determine

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

Explanation of Solution

Given: $h\left(x\right)={\left(x+1\right)}^5-5x-2$

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

  

Figure (1)

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