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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) -3
(B) 2
(C) 1
(D) -2

Let h(x)=2x37 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) 3-3\newline(B) 22\newline(C) 11\newline(D) 2-2

Full solution

Q. Let h(x)=2x37 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) 3-3\newline(B) 22\newline(C) 11\newline(D) 2-2
  1. Find Derivative: To find the absolute maximum value of the function h(x)h(x) on 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 where the derivative h(x)h'(x) is zero or undefined.
  2. Find Critical Points: First, we find the derivative of h(x)h(x). The derivative of h(x)=2x37h(x)=-2x^3-7 with respect to xx is h(x)=6x2h'(x)=-6x^2.
  3. Evaluate Function: Next, we set the derivative equal to zero to find the critical points: 6x2=0-6x^2 = 0. Solving for xx gives us x=0x = 0 as the only critical point within the interval [3,2][-3,2].
  4. Evaluate at 3-3: Now we evaluate the function h(x)h(x) at the critical point and the endpoints of the interval. We have three points to consider: x=3x = -3, x=0x = 0, and x=2x = 2.
  5. Evaluate at 00: Evaluating h(x)h(x) at x=3x = -3: h(3)=2(3)37=2(27)7=547=47h(-3) = -2(-3)^3 - 7 = -2(-27) - 7 = 54 - 7 = 47.
  6. Evaluate at 22: Evaluating h(x)h(x) at x=0x = 0: h(0)=2(0)37=7h(0) = -2(0)^3 - 7 = -7.
  7. Compare Values: Evaluating h(x)h(x) at x=2x = 2: h(2)=2(2)37=2(8)7=167=23h(2) = -2(2)^3 - 7 = -2(8) - 7 = -16 - 7 = -23.
  8. Maximum Value: Comparing the values of h(x)h(x) at x=3x = -3, x=0x = 0, and x=2x = 2, we find that the maximum value is 4747, which occurs at x=3x = -3.

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