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 =
4 answers
correction the function is e^((x)^(2) -4)
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)
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)
Thanks but I'm confused, what would the maximum be if the function if h(x)=e^((x)^(2) -4)
I Get it Now, thank you. This website is amazing.
0 answers