Chapter 4.2, Problem 59E

a)

To determine

To estimate: The absolute maximum and minimum values of the function to two decimal places.

Answer to Problem 59E

The absolute maximum value of $f\left(x\right)$ is $f\left(\frac{3}{4}\right)\approx 0.325$.

The absolute minimum value of $f\left(x\right)$ are $f\left(0\right)=0$ and $f\left(1\right)=0$.

Explanation of Solution

Given:

The function $f\left(x\right)=x\sqrt{x-x^2}$.

Graph:

Use the graphing calculator to draw the graph of the function $f\left(x\right)=x\sqrt{x-x^2}$ as shown below in Figure 1.

Calculation:

From the figure 1, it is observed that $f\left(0.75\right)\ge f\left(x\right)$ for all values of x in a domain. Therefore, the absolute maximum value of $f\left(x\right)$ occurs at $x=0.75$ and the value is $0.32$

Since $f\left(0\right)\le f\left(x\right)$ for all values of x in a domain. Therefore, the absolute minimum value of $f\left(x\right)$ occurs at $x=0,1$ and the values are $f\left(0\right)=0$ and $f\left(1\right)=0$.

Thus, the absolute maximum value of $f\left(x\right)$ is $f\left(0.75\right)\approx 0.32$ and the absolute minimum value of $f\left(x\right)$ is $f\left(0\right)=f\left(1\right)=0$.

(b)

To determine

To find: The exact maximum and minimum values of the given function $f\left(x\right)=x\sqrt{x-x^2}$.

Answer to Problem 59E

The absolute maximum value of $f\left(x\right)$ is

The absolute minimum value of $f\left(x\right)$ is are $f\left(0\right)=0$ and $f\left(1\right)=0$

Explanation of Solution

Obtain the first derivative of the given function.

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

Differentiate with respect to x.

$\begin{array}{c}{f}^{\prime }\left(x\right)=\frac{d}{dx}\left(x\sqrt{x-x^2}\right)\\ =\sqrt{x-x^2}\frac{d}{dx}\left(x\right)+x\frac{d}{dx}{\left(x-x^2\right)}^{\frac{1}{2}}\\ =\sqrt{x-x^2}\left(1\right)+x\frac{1}{2}{\left(x-x^2\right)}^{\frac{1}{2}-1}\frac{d}{dx}\left(x-x^2\right)\\ =\sqrt{x-x^2}+\frac{1}{2}x{\left(x-x^2\right)}^{\frac{-1}{2}}\left(1-2x\right)\end{array}$

On further simplification,

$\begin{array}{c}{f}^{\prime }\left(x\right)=\sqrt{x-x^2}+x\cdot \frac{1-2x}{2\sqrt{x-x^2}}\\ =\frac{2\sqrt{x-x^2}\sqrt{x-x^2}+x-2x^2}{2\sqrt{x-x^2}}\\ =\frac{2\left(x-x^2\right)+x-2x^2}{2\sqrt{x-x^2}}\\ =\frac{2x-2x^2+x-2x^2}{2\sqrt{x-x^2}}\end{array}$

$=\frac{3x-4x^2}{2\sqrt{x-x^2}}$

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

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

This gives $x=0,x=\frac{3}{4}$.

Here, the critical numbers are $x=0,x=\frac{3}{4}$.

The domain of $f$ is $0\le x\le 1$.

Hence all the critical numbers lie in the given interval $0\le x\le 1$.

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

When $x=0$,

$\begin{array}{c}f\left(0\right)=\left(0\right)\sqrt{0-{\left(0\right)}^2}\\ =0\end{array}$

When $x=1$,

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

When $x=\frac{3}{4}$,

$\begin{array}{c}f\left(\frac{3}{4}\right)=\left(\frac{3}{4}\right)\sqrt{\left(\frac{3}{4}\right)-{\left(\frac{3}{4}\right)}^2}\\ =\left(\frac{3}{4}\right)\sqrt{\left(\frac{3}{4}\right)-\left(\frac{9}{16}\right)}\\ =\frac{3}{4}\sqrt{\frac{3}{16}}\\ =\frac{3\sqrt{3}}{16}\end{array}$

$\approx 0.325$

Recall the fact that the largest functional value is absolute maximum and smallest functional value is absolute minimum.

Thus, the absolute maximum value of $f\left(x\right)$ is $f\left(\frac{3}{4}\right)\approx 0.325$ and the absolute minimum value of $f\left(x\right)$ are $f\left(0\right)=0$ and $f\left(1\right)=0$.