Chapter 4.3, Problem 26E

(a)

To determine

To find:The intervals of increase or decrease.

Explanation of Solution

Given: $h\left(x\right)=x^5-2x^3+x$

Calculate the first derivative of the given function.

  $\begin{array}{c}{h}^{\prime }\left(x\right)=5x^4-6x^2+1\\ 0=5x^4-6x^2+1\\ \left(x-1\right)\left(x+1\right)\left(5x^2-1\right)=0\\ x=\pm 1,\pm \frac{1}{\sqrt{5}}\end{array}$

The function is increasing in the interval $\left(-\infty ,-1\right)$

The function is decreasing in the interval $\left(-1,\frac{-1}{\sqrt{5}}\right)$ .

The function is increasing in the interval $\left(-\frac{1}{\sqrt{5}},\frac{1}{\sqrt{5}}\right)$

The function is decreasing in the interval $\left(\frac{1}{\sqrt{5}},1\right)$ .

The function is increasing in the interval $\left(1,\infty \right)$

(b)

To determine

To find :The local maxima and local minima.

Explanation of Solution

The function is increasing in the interval $\left(-\infty ,-1\right)$

The function is decreasing in the interval $\left(-1,\frac{-1}{\sqrt{5}}\right)$ .

The function is increasing in the interval $\left(-\frac{1}{\sqrt{5}},\frac{1}{\sqrt{5}}\right)$

The function is decreasing in the interval $\left(\frac{1}{\sqrt{5}},1\right)$ .

The function is increasing in the interval $\left(1,\infty \right)$

The value of $h^{\prime }$ is changing from positive to negative at $x=-1$ and $x=\frac{1}{\sqrt{5}}$

So, the local maxima is $\left(-1,0\right)\text{and}\left(\frac{1}{\sqrt{5}},0.286\right)$

The value of $f^{\prime }$ is changing from negative to positive at $x=-\frac{1}{\sqrt{5}}$ and $x=1$ .

So, the local minima is $\left(-\frac{1}{\sqrt{5}},-0.286\right)\text{and}\left(1,0\right)$

(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.

  $\begin{array}{c}{h}^{″}\left(x\right)=20x^3-12x\\ 0=20x^3-12x\\ 20x^3-12x\\ x=\pm 0.775\end{array}$

  $f^{″}\left(x\right)<0$ for the interval $x<-0.775$ . So, the graph is concave down for $\left(-\infty ,-0.775\right)$ .

  $f^{″}\left(x\right)>0$ for the interval $-0.775<x<0$ . So, the graph is concave up for $\left(-0.775,0\right)$

  $f^{″}\left(x\right)<0$ for the interval $0<x<0.775$ . So, the graph is concave down for $\left(0,0.775\right)$ .

  $f^{″}\left(x\right)>0$ for the interval $x>0.775$ . So, the graph is concave up for $\left(0.775,\infty \right)$

The inflection point is $\left(-0.775,-0.123\right),\left(0,0\right)\text{and}\left(0.775,0.123\right)$ .

(d)

To determine

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

Explanation of Solution

Given: $h\left(x\right)=x^5-2x^3+x$

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

  

Figure (1)

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