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 =

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)

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.

To find the absolute extrema of a function, you need to evaluate the function at both the critical points and endpoints of the interval. Here are the steps to follow:

1. Find the critical points by taking the derivative of the function and setting it equal to zero.

2. Evaluate the function at each critical point and the endpoints of the interval.

3. Compare the function values to determine the absolute maximum and minimum.

Now let's solve the problem step by step:

Step 1: Find the critical points.
To find the critical points, we need to take the derivative of the function.
h'(x) = 2x * e^x^2

To find where the derivative is equal to zero, set h'(x) = 0 and solve for x:
2x * e^x^2 = 0

Since e^x^2 is always positive, the derivative can only be zero if 2x = 0.
Solving for x, we get x = 0.

So, the only critical point in the interval [-2, 2] is x = 0.

Step 2: Evaluate the function at the critical point and endpoints.
Evaluate h(x) at x = -2, 0, and 2.

h(-2) = e^(-2^2) - 4 = e^4 - 4
h(0) = e^(0^2) - 4 = e^0 - 4 = 1 - 4 = -3
h(2) = e^(2^2) - 4 = e^4 - 4

Step 3: Compare the function values.
Comparing the function values, we can find the absolute maximum and minimum.

The absolute maximum value is the largest value among h(-2), h(0), and h(2).
The absolute minimum value is the smallest value among h(-2), h(0), and h(2).

Comparing the values, we can see that h(-2) = e^4 - 4 is the largest value, so the absolute maximum value occurs at x = -2.

Similarly, h(0) = -3 is the smallest value, so the absolute minimum value occurs at x = 0.

Therefore:
Absolute maximum value: at x = -2
Absolute minimum value: at x = 0