Chapter 4.2, Problem 57E

a)

To determine

To Estimate: The absolute maximum and minimum values (round off up to two decimals) of the function from its graph.

Answer to Problem 57E

The absolute maximum value of $f\left(x\right)$ is $f\left(-0.775\right)\approx 2.19$.

The absolute minimum value of $f\left(x\right)$ is $f\left(0.775\right)\approx 1.81$.

Explanation of Solution

Given:

The function $f\left(x\right)=x^5-x^3+2,-1\le x\le 1$.

Graph:

Use the online graphing calculator to draw the graph of the function $f\left(x\right)=x^5-x^3+2$ in the interval $-1\le x\le 1$ as shown below in Figure 1.

Calculation:

From Figure 1, it is observed that at the point $\left(-0.775,2.186\right)$ the curve attains the maximum value. Therefore, the absolute maximum value occurs at $x=-0.775$ and the value is $2.186$.

Also from Figure 1, it is observed that at the point $\left(0.775,1.814\right)$ the curve attains the minimum value. Therefore, the absolute minimum value occurs at $x=0.775$ and the value is $1.814$.

Thus, the absolute maximum value of $f\left(x\right)$ is $f\left(-0.775\right)\approx 2.19$ and the absolute minimum value of $f\left(x\right)$ is $f\left(0.775\right)\approx 1.81$.

b)

To determine

To find: The exact maximum and minimum values of the function $f\left(x\right)=x^5-x^3+2,-1\le x\le 1$.

Answer to Problem 57E

The absolute maximum value of $f\left(x\right)$ is $f\left(-\sqrt{\frac{3}{5}}\right)\approx 2.1859$

The absolute minimum value of $f\left(x\right)$ is $f\left(\sqrt{\frac{3}{5}}\right)\approx 1.8141$

Explanation of Solution

Obtain the first derivative of the given function.

$f^{\prime }\left(x\right)=\frac{d}{dx}\left(x^5-x^3+2\right)$

Differentiate with respect to x.

$f^{\prime }\left(x\right)=5x^4-3x^2$

Set $f^{\prime }\left(x\right)=0$ and obtain the critical point.

$\begin{array}{l}5x^4-3x^2=0\\ x^2\left(5x^2-3\right)=0\\ x^2=0,5x^2-3=0\\ x=0,x=\pm \sqrt{\frac{3}{5}}\end{array}$

Here, the critical numbers are $x=-\sqrt{\frac{3}{5}},0,\sqrt{\frac{3}{5}}$.

All of the critical numbers lie in the given interval $-1\le x\le 1$.

Apply the extreme values of the given interval and the critical number values in $f\left(x\right)$ .

When $x=-1$,

$\begin{array}{c}f\left(-1\right)={\left(-1\right)}^5-{\left(-1\right)}^3+2\\ =-1+1+2\\ =2\end{array}$

When $x=1$,

$\begin{array}{c}f\left(1\right)={\left(1\right)}^5-{\left(1\right)}^3+2\\ =1-1+2\\ =2\end{array}$

When $x=0$,

$\begin{array}{c}f\left(1\right)={\left(0\right)}^5-{\left(0\right)}^3+2\\ =2\end{array}$

When $x=-\sqrt{\frac{3}{5}}$

$\begin{array}{c}f\left(-\sqrt{\frac{3}{5}}\right)={\left(-\sqrt{\frac{3}{5}}\right)}^5-{\left(-\sqrt{\frac{3}{5}}\right)}^3+2\\ =-\frac{9\sqrt{3}}{25\sqrt{5}}+\frac{3\sqrt{3}}{5\sqrt{5}}+2\\ =\sqrt{\frac{3}{5}}\left[\frac{9}{25}+\frac{3}{5}\right]+2\\ =\frac{6}{25}\sqrt{\frac{3}{5}}+2\end{array}$

$\approx 2.1859$

When $x=\sqrt{\frac{3}{5}}$

$\begin{array}{c}f\left(\sqrt{\frac{3}{5}}\right)={\left(\sqrt{\frac{3}{5}}\right)}^5-{\left(\sqrt{\frac{3}{5}}\right)}^3+2\\ =\frac{9\sqrt{3}}{25\sqrt{5}}-\frac{3\sqrt{3}}{5\sqrt{5}}+2\\ =\sqrt{\frac{3}{5}}\left[\frac{9}{25}-\frac{3}{5}\right]+2\\ =-\frac{6}{25}\sqrt{\frac{3}{5}}+2\end{array}$

$\approx 1.8141$

Recall the fact that the largest functional value is absolute maximum and smallest functional value is absolute minimum. Compare with above values.

Thus, the absolute maximum value of $f\left(x\right)$ is $f\left(-\sqrt{\frac{3}{5}}\right)\approx 2.1859$ and the absolute minimum value of $f\left(x\right)$ is $f\left(\sqrt{\frac{3}{5}}\right)\approx 1.8141$.