Chapter 8.3, Problem 17Q

To determine

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

Answer to Problem 17Q

The function $f\left(x\right)$ is maximum at $x=-1$ . The maximum value is $f\left(-1\right)=-8$

Explanation of Solution

Given information:

The function is,

  $f\left(x\right)=5x^2+10x-3$

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:

  $f\left(x\right)=5x^2+10x-3$

Differentiate with respect to x,

  $\begin{array}{l}f\text{'}\left(x\right)=\frac{d}{dx}\left(5x^2+10x-3\right)\\ f\text{'}\left(x\right)=5\frac{d}{dx}\left(x^2\right)+10\frac{d}{dx}\left(x\right)-\frac{d}{dx}\left(3\right)\\ f\text{'}\left(x\right)=5\left(2x^{2-1}\right)+10\left(1\right)-\left(0\right)\left[\because \frac{d}{dv}\left(v^n\right)=n{v}^{n-1}\text{and}\frac{d}{dv}\left(\text{constant}\right)=0\right]\\ f\text{'}\left(x\right)=10x+10\end{array}$

For critical point,

  $\begin{array}{c}f\text{'}\left(x\right)=0\\ 10x+10=0\end{array}$

Subtract 10 from both sides,

  $\begin{array}{c}10x+10-10=-10\\ 10x=-10\end{array}$

Divide by 10,

  $\begin{array}{c}\frac{10x}{10}=-\frac{10}{10}\\ x=-1\end{array}$

Critical point is, $x=-1$

Again, differentiate $f\text{'}\left(x\right)=10x+10$ ,

  $\begin{array}{l}f\text{'}\text{'}\left(x\right)=\frac{d}{dx}\left(10x+10\right)\\ f\text{'}\text{'}\left(x\right)=10\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(10\right)\\ f\text{'}\text{'}\left(x\right)=10\left(1\right)+\left(0\right)\left[\because \frac{d}{dv}\left(v^n\right)=n{v}^{n-1}\text{and}\frac{d}{dv}\left(\text{constant}\right)=0\right]\\ f\text{'}\text{'}\left(x\right)=10\\ f\text{'}\text{'}\left(x\right)>0\end{array}$

Second derivative is positive at point $x=-1$ .

So, the function $f\left(x\right)$ is minimum at $x=-1$ .

For maximum value, substitute $x=-1$ in function $f\left(x\right)$ .

  $\begin{array}{c}f\left(x\right)=5x^2+10x-3\\ f\left(-1\right)=5{\left(-1\right)}^2+10\left(-1\right)-3\\ f\left(-1\right)=5\left(1\right)-10-3\\ f\left(-1\right)=5-10-3\\ f\left(-1\right)=-8\end{array}$

So, the function $f\left(x\right)$ is maximum at $x=-1$ . The maximum value is $f\left(-1\right)=-8$