Chapter 8.3, Problem 20Q

To determine

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

Answer to Problem 20Q

The function $f\left(x\right)$ is minimum at $x=-2$ . The minimum value is $f\left(x\right)=-5$ .

Explanation of Solution

Given information:

The function is,

  $f\left(x\right)=2x^2+8x+3$

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:

  $f\left(x\right)=2x^2+8x+3$

Differentiate with respect to x,

  $\begin{array}{l}f\text{'}\left(x\right)=\frac{d}{dx}\left(2x^2+8x+3\right)\\ f\text{'}\left(x\right)=2\frac{d}{dx}\left(x^2\right)+8\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(3\right)\\ f\text{'}\left(x\right)=2\left(2x^{2-1}\right)+8\left(1\right)+\left(0\right)\left[\because \frac{d}{dv}\left(v^n\right)=n{v}^{n-1}\text{and}\frac{d}{dv}\left(\text{constant}\right)=0\right]\\ f\text{'}\left(x\right)=4x+8\end{array}$

For critical point,

  $\begin{array}{l}f\text{'}\left(x\right)=0\\ 4x+8=0\end{array}$

Subtract 8 from both sides,

  $\begin{array}{c}4x+8-8=-8\\ 4x=-8\end{array}$

Divide by 4 from both sides,

  $\begin{array}{c}\frac{4x}{4}=-\frac{8}{4}\\ x=-2\end{array}$

Critical point is, $x=-2$

Again, differentiate $f\text{'}\left(x\right)=4x+8$ ,

  $\begin{array}{l}f\text{'}\text{'}\left(x\right)=\frac{d}{dx}\left(4x+8\right)\\ f\text{'}\text{'}\left(x\right)=4\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(8\right)\\ f\text{'}\text{'}\left(x\right)=4\left(1\right)+\left(0\right)\left[\because \frac{d}{dv}\left(v^n\right)=n{v}^{n-1}\text{and}\frac{d}{dv}\left(\text{constant}\right)=0\right]\\ f\text{'}\text{'}\left(x\right)=4\\ f\text{'}\text{'}\left(x\right)>0\end{array}$

Second derivative is positive at point $x=-2$ .

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

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

  $\begin{array}{l}f\left(x\right)=2x^2+8x+3\\ f\left(x\right)=2{\left(-2\right)}^2+8\left(-2\right)+3\\ f\left(x\right)=2\left(4\right)-16+3\\ f\left(x\right)=8-16+3\\ f\left(x\right)=-5\end{array}$

So, the function $f\left(x\right)$ is minimum at $x=-2$ . The minimum value is $f\left(x\right)=-5$ .