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Identify the absolute minimum value of h(x)=−2x2−1 over [−4,3].

User Gazreese
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To find the absolute minimum value of the function h(x) = -2x^2 - 1 over the interval [-4, 3], we need to evaluate the function at both endpoints of the interval and at any critical points within the interval.

1. Evaluate the function at the endpoints of the interval:

h(-4) = -2(-4)^2 - 1 = -2(16) - 1 = -32 - 1 = -33

h(3) = -2(3)^2 - 1 = -2(9) - 1 = -18 - 1 = -19

2. Find the critical points (where the derivative of the function is equal to zero or undefined) and evaluate the function at those points. Taking the derivative of h(x) and setting it to zero, we get:

h'(x) = -4x

Setting h'(x) = 0, we find the critical point: x = 0.

Evaluate h(x) at x = 0:

h(0) = -2(0)^2 - 1 = -2(0) - 1 = -1

Now, compare the values at the endpoints and the critical point to determine the absolute minimum:

The absolute minimum value of h(x) = -2x^2 - 1 over the interval [-4, 3] is -33, which occurs at x = -4.

User Jay Jargot
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