Chapter 4.1, Problem 14E

To determine

To find : extreme values of the function on the interval using the Candidates test.

Answer to Problem 14E

Maximum occur at $\left(0,1\right)$ and minimum occur at $\left(-2,e^{-4}\right)$ and, $\left(3,e^{-9}\right)$ .

Explanation of Solution

Given information :

  $k\left(x\right)=e^{-x^2}$ , $-2\le x\le 3$

Formula needed :

For critical points, $\frac{dy}{dx}=0$ or $\frac{dy}{dx}$ is not defined.

Differentiate $k\left(x\right)=e^{-x^2}$ with respect to $x$ .

  $\begin{array}{c}{k}^{\prime }\left(x\right)=\frac{d}{dx}{e}^{-x^2}\\ =e^{-x^2}\cdot \frac{d}{dx}\left(-x^2\right)\\ =e^{-x^2}\cdot \left(-2x\right)\\ =-2x{e}^{-x^2}\end{array}$

Take $\frac{dy}{dx}=0$ .

  $\begin{array}{c}-2x{e}^{-x^2}=0\\ x=0\end{array}$

One critical point exists in the given interval.

Plug the critical point in the function.

  $\begin{array}{c}k\left(0\right)=e^{-0^2}\\ =e^0\\ =1\end{array}$

Plug the end points in the function.

  $\begin{array}{c}k\left(-2\right)=e^{-{\left(-2\right)}^2}\\ =e^{-4}\end{array}$

And

  $\begin{array}{c}k\left(3\right)=e^{-{\left(3\right)}^2}\\ =e^{-9}\end{array}$

Maximum occur at $\left(0,1\right)$ and minimum occur at $\left(-2,e^{-4}\right)$ and, $\left(3,e^{-9}\right)$ .