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Let h(x)=-2x^(3)-7. The absolute maximum value of h over the closed interval [-3,2] occurs at what x-value? Choose 1 answer: (A) 1 (B) -2 (C) 2 (D) -3

Let h(x)=−2x3−7 h(x)=-2 x^{3}-7 .\newlineThe absolute maximum value of h h over the closed interval [−3,2] [-3,2] occurs at what x x -value?\newlineChoose 11 answer:\newline(A) 11\newline(B) −2-2\newline(C) 22\newline(D) −3-3

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Q. Let h(x)=−2x3−7 h(x)=-2 x^{3}-7 .\newlineThe absolute maximum value of h h over the closed interval [−3,2] [-3,2] occurs at what x x -value?\newlineChoose 11 answer:\newline(A) 11\newline(B) −2-2\newline(C) 22\newline(D) −3-3
  1. Find Derivative of h(x): To find the absolute maximum value of the function h(x)h(x) over the closed interval [−3,2][-3,2], we need to evaluate the function at the critical points and the endpoints of the interval. The critical points are found where the derivative h′(x)h'(x) is zero or undefined.\newlineLet's find the derivative of h(x)h(x).\newlineh′(x)=ddx(−2x3−7)h'(x) = \frac{d}{dx} (-2x^3 - 7)\newlineh′(x)=−6x2h'(x) = -6x^2
  2. Find Critical Points: Now we need to find the critical points by setting the derivative equal to zero and solving for xx.
    −6x2=0-6x^2 = 0
    x2=0x^2 = 0
    x=0x = 0
    The only critical point in the interval [−3,2][-3,2] is x=0x = 0.
  3. Evaluate Function: Next, we evaluate the function h(x)h(x) at the critical point and the endpoints of the interval.h(−3)=−2(−3)3−7=−2(−27)−7=54−7=47h(-3) = -2(-3)^3 - 7 = -2(-27) - 7 = 54 - 7 = 47h(0)=−2(0)3−7=−7h(0) = -2(0)^3 - 7 = -7h(2)=−2(2)3−7=−2(8)−7=−16−7=−23h(2) = -2(2)^3 - 7 = -2(8) - 7 = -16 - 7 = -23
  4. Compare Values: We compare the values of h(x)h(x) at x=−3x = -3, x=0x = 0, and x=2x = 2 to find the absolute maximum.\newlineh(−3)=47h(-3) = 47\newlineh(0)=−7h(0) = -7\newlineh(2)=−23h(2) = -23\newlineThe largest value is h(−3)=47h(-3) = 47, so the absolute maximum occurs at x=−3x = -3.

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