Question

find the absolute maximum and minimum values of the following function on the given interval. then graph the function g(x)=e^{-x^{2}},-3leq xleq2
find the absolute maximum. select the correct choice below and, if necessary, fill in the answer box to complete your choice
the absolute maximum value 1 occurs at x = 0. (type exact answers. use a comma to separate answers as needed ) there is no absolute maximum.
find the absolute minimum. select the correct choice below and, if necessary, fill in the answer box to complete your choice
a. the absolute minimum value occurs at x = (type exact answers. use a comma to separate answers as needed )
b. there is no absolute minimum.

Explanation:

Step1: Find the derivative

Let $u = -x^{2}$, then $g(x)=e^{u}$. By the chain - rule $\frac{dg}{dx}=\frac{dg}{du}\cdot\frac{du}{dx}$. We know that $\frac{d}{du}(e^{u}) = e^{u}$ and $\frac{d}{dx}(-x^{2})=-2x$. So $g'(x)=e^{-x^{2}}\cdot(-2x)=-2xe^{-x^{2}}$.

Step2: Find the critical points

Set $g'(x) = 0$. Since $e^{-x^{2}}\gt0$ for all real $x$, then $-2xe^{-x^{2}} = 0$ when $x = 0$.

Step3: Evaluate the function at critical points and endpoints

Evaluate $g(x)$ at $x=-3,x = 0,x = 2$.
When $x=-3$, $g(-3)=e^{-(-3)^{2}}=e^{-9}=\frac{1}{e^{9}}$.
When $x = 0$, $g(0)=e^{-0^{2}}=1$.
When $x = 2$, $g(2)=e^{-2^{2}}=e^{-4}=\frac{1}{e^{4}}$.
Since $\frac{1}{e^{9}}\lt\frac{1}{e^{4}}\lt1$.

Answer:

The absolute minimum value $\frac{1}{e^{9}}$ occurs at $x=-3$.