Question

Im having a real difficult time solving this question.

Find the absolute extrema of the function.
h(x) = e^x^(2) - 4 on [-2,2]
Absolute maximum value:
at x =

Absolute minimum value:
at x =

Answer 1

correction the function is e^((x)^(2) -4)

Answer 2

well, x^2-4 has a minimum at x=0, so e^(x^2-4) will as well.

since x^2-4 has an absolute max at x=±2, so does e^(x^2-4)

But, as long as this is calculus,

dh/dx = 2x e^(x^2-4)
dh/dx=0 at x=0 (minimum)

Answer 3

Thanks but I'm confused, what would the maximum be if the function if h(x)=e^((x)^(2) -4)

Answer 4

I Get it Now, thank you. This website is amazing.