Q. Let h(x)=−2x3−7.The absolute maximum value of h over the closed interval [−3,2] occurs at what x-value?Choose 1 answer:(A) −3(B) 2(C) 1(D) −2
Find Derivative: To find the absolute maximum value of the function h(x) on the closed interval [−3,2], we need to evaluate the function at the critical points and the endpoints of the interval. The critical points are where the derivative h′(x) is zero or undefined.
Find Critical Points: First, we find the derivative of h(x). The derivative of h(x)=−2x3−7 with respect to x is h′(x)=−6x2.
Evaluate Function: Next, we set the derivative equal to zero to find the critical points: −6x2=0. Solving for x gives us x=0 as the only critical point within the interval [−3,2].
Evaluate at −3: Now we evaluate the function h(x) at the critical point and the endpoints of the interval. We have three points to consider: x=−3, x=0, and x=2.
Evaluate at 0: Evaluating h(x) at x=−3: h(−3)=−2(−3)3−7=−2(−27)−7=54−7=47.
Evaluate at 2: Evaluating h(x) at x=0: h(0)=−2(0)3−7=−7.
Compare Values: Evaluating h(x) at x=2: h(2)=−2(2)3−7=−2(8)−7=−16−7=−23.
Maximum Value: Comparing the values of h(x) at x=−3, x=0, and x=2, we find that the maximum value is 47, which occurs at x=−3.