Chapter 8.3, Problem 18Q

To determine

To tell whether the function has a minimum value or a maximum value, and then find the value.

Answer to Problem 18Q

The function $f\left(x\right)$ is maximum at $x=2$ . The maximum value is $f\left(2\right)=18$

Explanation of Solution

Given information:

The function is,

  $f\left(x\right)=-\frac{1}{2}{x}^2+2x+16$

Second derivative test:

If a function has a critical point for which $f\text{'}\left(x\right)=0$ and the second derivative is positive at this point, then $f$ is minimum.

If a function has a critical point for which $f\text{'}\left(x\right)=0$ and the second derivative is negative at this point, then $f$ is maximum.

Calculation:

  $f\left(x\right)=-\frac{1}{2}{x}^2+2x+16$

Differentiate with respect to x,

  $\begin{array}{l}f\text{'}\left(x\right)=\frac{d}{dx}\left(-\frac{1}{2}{x}^2+2x+16\right)\\ f\text{'}\left(x\right)=-\frac{1}{2}\frac{d}{dx}\left(x^2\right)+2\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(16\right)\\ f\text{'}\left(x\right)=-\frac{1}{2}\left(2x^{2-1}\right)+2\left(1\right)+\left(0\right)\left[\because \frac{d}{dv}\left(v^n\right)=n{v}^{n-1}\text{and}\frac{d}{dv}\left(\text{constant}\right)=0\right]\\ f\text{'}\left(x\right)=-\frac{2x}{2}+2\\ f\text{'}\left(x\right)=-x+2\end{array}$

For critical point,

  $\begin{array}{c}f\text{'}\left(x\right)=0\\ -x+2=0\end{array}$

Add x from both sides,

  $\begin{array}{c}x-x+2=x\\ 2=x\end{array}$

Critical point is, $x=2$

Again, differentiate $f\text{'}\left(x\right)=-x+2$ ,

  $\begin{array}{l}f\text{'}\text{'}\left(x\right)=\frac{d}{dx}\left(-x+2\right)\\ f\text{'}\text{'}\left(x\right)=-\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(2\right)\\ f\text{'}\text{'}\left(x\right)=-\left(1\right)+\left(0\right)\left[\because \frac{d}{dv}\left(v^n\right)=n{v}^{n-1}\text{and}\frac{d}{dv}\left(\text{constant}\right)=0\right]\\ f\text{'}\text{'}\left(x\right)=-1\\ f\text{'}\text{'}\left(x\right)<0\end{array}$

Second derivative is negative at point $x=2$ .

So, the function $f\left(x\right)$ is maximum at $x=2$ .

For maximum value, substitute $x=2$ in function $f\left(x\right)$ .

  $\begin{array}{l}f\left(x\right)=-\frac{1}{2}{x}^2+2x+16\\ f\left(2\right)=-\frac{1}{2}{\left(2\right)}^2+2\left(2\right)+16\\ f\left(2\right)=-\frac{4}{2}+4+16\\ f\left(2\right)=-2+4+16\\ f\left(2\right)=18\end{array}$

So, the function $f\left(x\right)$ is maximum at $x=2$ . The maximum value is $f\left(2\right)=18$ .