Chapter 4.2, Problem 23E

a.

To determine

To find: the local extrema of the given function.

Answer to Problem 23E

The local maximum point is $\left(\frac{8}{3},\frac{16\sqrt{3}}{9}\right)$ and there is no local minimum.

Explanation of Solution

Given:

  $f\left(x\right)=x\sqrt{4-x}$ .

Calculation:

Consider the given function,

  $f\left(x\right)=x\sqrt{4-x}$

Differentiating with respect to $x$ , it follows that

  $\begin{array}{c}{f}^{\prime }\left(x\right)=1\cdot \sqrt{4-x}+\frac{1}{2\sqrt{4-x}}\left(0-1\right)\cdot x\\ =\sqrt{4-x}-\frac{x}{2\sqrt{4-x}}\\ =\frac{2\left(4-x\right)-x}{2\sqrt{4-x}}\\ =\frac{8-2x-x}{2\sqrt{4-x}}\\ =\frac{8-3x}{2\sqrt{4-x}}\end{array}$

Now to find the critical values, solve the first derivative by equating it to zero. That is,

  $\begin{array}{c}{f}^{\prime }\left(x\right)=0\\ \frac{8-3x}{2\sqrt{4-x}}=0\\ 8-3x=0\\ 3x=8\\ x=\frac{8}{3}\end{array}$

Therefore $x=\frac{8}{3}$ is the critical value.

The graph of $f\left(x\right)=x\sqrt{4-x}$ rises to the left of $x=\frac{8}{3}$ and falls to the right of $x=\frac{8}{3}$ .

Therefore $x=\frac{8}{3}$ is the point of local maxima.

Now the local maximum value is,

  $\begin{array}{c}f\left(\frac{8}{3}\right)=\frac{8}{3}\sqrt{4-\frac{8}{3}}\\ =\frac{8}{3}\sqrt{\frac{4}{3}}\\ =\frac{16}{3\sqrt{3}}\\ =\frac{16\sqrt{3}}{9}\end{array}$

b.

To determine

To find: the intervals in which the given function is increasing.

Answer to Problem 23E

The given function increases in the interval $\left(-\infty ,\frac{8}{3}\right)$ .

Explanation of Solution

Given:

  $f\left(x\right)=x\sqrt{4-x}$ .

Concept used:

The function increases in the interval at which the first derivative is positive.

Calculation:

Consider the given function,

  $f\left(x\right)=x\sqrt{4-x}$

Differentiating with respect to $x$ , it follows that

  $\begin{array}{c}{f}^{\prime }\left(x\right)=1\cdot \sqrt{4-x}+\frac{1}{2\sqrt{4-x}}\left(0-1\right)\cdot x\\ =\sqrt{4-x}-\frac{x}{2\sqrt{4-x}}\\ =\frac{2\left(4-x\right)-x}{2\sqrt{4-x}}\\ =\frac{8-2x-x}{2\sqrt{4-x}}\\ =\frac{8-3x}{2\sqrt{4-x}}\end{array}$

Now the function increases in the interval for which,

  $\begin{array}{c}{f}^{\prime }\left(x\right)>0\\ \frac{8-3x}{2\sqrt{4-x}}>0\\ 8-3x>0\\ 3x<8\\ x<\frac{8}{3}\end{array}$

Hence the given function increases in the interval $\left(-\infty ,\frac{8}{3}\right)$ .

c.

To determine

To find: the intervals in which the given function is decreasing.

Answer to Problem 23E

The given function decreases in the interval $\left(\frac{8}{3},\infty \right)$ .

Explanation of Solution

Given:

  $f\left(x\right)=x\sqrt{4-x}$ .

Concept used:

The function decreases in the interval at which the first derivative is negative.

Calculation:

Consider the given function,

  $f\left(x\right)=x\sqrt{4-x}$

Differentiating with respect to $x$ , it follows that

  $\begin{array}{c}{f}^{\prime }\left(x\right)=1\cdot \sqrt{4-x}+\frac{1}{2\sqrt{4-x}}\left(0-1\right)\cdot x\\ =\sqrt{4-x}-\frac{x}{2\sqrt{4-x}}\\ =\frac{2\left(4-x\right)-x}{2\sqrt{4-x}}\\ =\frac{8-2x-x}{2\sqrt{4-x}}\\ =\frac{8-3x}{2\sqrt{4-x}}\end{array}$

Now the function decreases in the interval for which,

  $\begin{array}{c}{f}^{\prime }\left(x\right)<0\\ \frac{8-3x}{2\sqrt{4-x}}<0\\ 8-3x<0\\ 3x>8\\ x>\frac{8}{3}\end{array}$

Hence the given function decreases in the interval $\left(\frac{8}{3},\infty \right)$ .