Chapter 4.2, Problem 15E

a.

To determine

To find: the local extrema of the given function.

Answer to Problem 15E

The local maximum value is $\frac{25}{4}$ and there is no local minimum.

Explanation of Solution

Given:

  $f\left(x\right)=5x-x^2$ .

Calculation:

Consider the given function,

  $f\left(x\right)=5x-x^2$

Differentiating with respect to $x$ , it follows that

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

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

  $\begin{array}{c}{f}^{\prime }\left(x\right)=0\\ 5-2x=0\\ 2x=5\\ x=\frac{5}{2}\end{array}$

Therefore $x=\frac{5}{2}$ is the critical value.

The graph of $f\left(x\right)=5x-x^2$ rises to the left of $x=\frac{5}{2}$ and falls to the right of $x=\frac{5}{2}$ .

Therefore $x=\frac{5}{2}$ is the point of local maxima.

Now the local maximum value is,

  $\begin{array}{c}f\left(\frac{5}{2}\right)=5\left(\frac{5}{2}\right)-{\left(\frac{5}{2}\right)}^2\\ =\frac{25}{2}-\frac{25}{4}\\ =\frac{25}{4}\end{array}$

b.

To determine

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

Answer to Problem 15E

The given function increases in the interval $\left(-\infty ,\frac{5}{2}\right)$ .

Explanation of Solution

Given:

  $f\left(x\right)=5x-x^2$ .

Concept used:

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

Calculation:

Consider the given function,

  $f\left(x\right)=5x-x^2$

Differentiating with respect to $x$ , it follows that

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

Now the function increases in the interval for which,

  $\begin{array}{c}{f}^{\prime }\left(x\right)>0\\ 5-2x>0\\ 2x<5\\ x<\frac{5}{2}\end{array}$

Hence the given function increases in the interval $\left(-\infty ,\frac{5}{2}\right)$ .

c.

To determine

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

Answer to Problem 15E

The given function decreases in the interval $\left(\frac{5}{2},\infty \right)$ .

Explanation of Solution

Given:

  $f\left(x\right)=5x-x^2$ .

Concept used:

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

Calculation:

Consider the given function,

  $f\left(x\right)=5x-x^2$

Differentiating with respect to $x$ , it follows that

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

Now the function decreases in the interval for which,

  $\begin{array}{c}{f}^{\prime }\left(x\right)<0\\ 5-2x<0\\ 2x>5\\ x>\frac{5}{2}\end{array}$

Hence the given function decreases in the interval $\left(\frac{5}{2},\infty \right)$ .