Find the absolute extrema of the function. h(x) = ex2 − 4 on [−2, 2] Absolute maximum value: at x = ± Absolute minimum value: at x =

Answer 1

Hello Bernadette,

It looks like h(x) = e^x2 - 4

Take its derivative: h'(x) = 2xe^x2

Find critical points by setting h'(x) = 0: 2xe^x2 = 0 when x = 0

Now you'll have to plug three x-values back into h(x): x = 0 (critical point), x = -2 and x = 2 (because they both are endpoints of the interval)

h(0) = e⁰ - 4 = 1 - 4 = -3

h(-2) = e⁴ - 4 = 50.598...

h(2) = e⁴ - 4 = 50.598...

As you can see, -3 is the lowest value of all three. e⁴ - 4 is the highest and it happens at BOTH endpoints.

Therefore: The absolute maximum value, which is e⁴ - 4, happens at x = ±2. The absolute minimum value, which is -3, happens at x = 0.