Chapter 4, Problem 2RE

To determine

To find: The local maximum and minimum and the absolute maximum and absolute minimum of the function $f\left(x\right)=x\sqrt{1-x}$ on the interval [−1, 1].

Answer to Problem 2RE

The local maximum occurs at $x=\frac{2}{3}$.

The absolute maxima occurs at $x=\frac{2}{3}$.

The absolute minimum is $x=-1$.

Explanation of Solution

Given:

The function is, $f\left(x\right)=x\sqrt{1-x}$ and the interval is, [−1, 1].

Calculation:

Obtain the first derivative of the given function.

$\begin{array}{c}f\text{'}\left(x\right)=\left(1\right)\sqrt{1-x}-\frac{x}{2\sqrt{1-x}}\\ =\frac{2\left(1-x\right)-x}{2\sqrt{1-x}}\\ =\frac{2-3x}{2\sqrt{1-x}}\end{array}$

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

$\begin{array}{l}\frac{2-3x}{2\sqrt{1-x}}=0\\ 2-3x=0\\ x=\frac{2}{3}\end{array}$

The critical number is $x=\frac{2}{3}$ which lies on the given interval $\left[-1,1\right]$.

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

Substitute $x=-1$ in $f\left(x\right)$,

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

Substitute $x=\frac{2}{3}$ in $f\left(x\right)$,

$\begin{array}{c}f\left(\frac{2}{3}\right)=\left(\frac{2}{3}\right)\sqrt{1-\left(\frac{2}{3}\right)}\\ =\frac{2}{3}\sqrt{\frac{1}{3}}\end{array}$

Substitute $x=1$ in $f\left(x\right)$,

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

Since the largest functional value is the absolute maximum and the smallest functional value is the absolute minimum, the absolute maximum of $f\left(x\right)$ is $\frac{2}{3}\sqrt{\frac{1}{3}}$ and the absolute minimum of $f\left(x\right)$ is $-\sqrt{2}$.

Therefore, the absolute maxima of $f\left(x\right)$ occurs at $x=\frac{2}{3}$ and the absolute minimum of $f\left(x\right)$ is occurs at $x=-1$.

To find the local maximum and minimum obtain the second derivative of the function.

$\begin{array}{c}f\text{'}\text{'}\left(x\right)=\frac{d}{dx}\left(\frac{2-3x}{2\sqrt{1-x}}\right)\\ =\frac{\left(2\sqrt{1-x}\right)\left(-3\right)-\left(2-3x\right)\left(\frac{-1}{2\left(1-x\right)\sqrt{1-x}}\right)}{{\left(2\sqrt{1-x}\right)}^2}\\ =\frac{\left(2\sqrt{1-x}\right)\left(-3\right)+\left(\frac{2-3x}{2\left(1-x\right)\sqrt{1-x}}\right)}{{\left(2\sqrt{1-x}\right)}^2}\end{array}$.

Substitute the critical numbers in the second derivative and obtain the local maximum and minimum as follows.

Substitute $x=\frac{2}{3}$ in $f\text{'}\text{'}\left(x\right)$,

$\begin{array}{c}f\text{'}\text{'}\left(\frac{2}{3}\right)=\frac{\left(2\sqrt{1-\frac{2}{3}}\right)\left(-3\right)+\left(\frac{2-3\cdot \frac{2}{3}}{2\left(1-\frac{2}{3}\right)\sqrt{1-\frac{2}{3}}}\right)}{{\left(2\sqrt{1-\frac{2}{3}}\right)}^2}\\ =\frac{\left(-6\sqrt{\frac{1}{3}}\right)+\left(\frac{2-2}{2\left(\frac{1}{3}\right)\sqrt{\frac{1}{3}}}\right)}{{\left(2\sqrt{\frac{1}{3}}\right)}^2}\\ =\frac{\left(-6\sqrt{\frac{1}{3}}\right)+0}{\frac{4}{3}}<0\end{array}$

Thus, the local maximum occurs at $x=\frac{2}{3}$.