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Let f(x)=-x^(3)+3x^(2)-6. The absolute maximum value of f over the closed interval [-2,5] occurs at what x-value? Choose 1 answer: (A) 5 (B) 0 (C) 2 (D) -2

Let f(x)=−x3+3x2−6 f(x)=-x^{3}+3 x^{2}-6 .\newlineThe absolute maximum value of f f over the closed interval [−2,5] [-2,5] occurs at what x x -value?\newlineChoose 11 answer:\newline(A) 55\newline(B) 00\newline(C) 22\newline(D) −2-2

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Q. Let f(x)=−x3+3x2−6 f(x)=-x^{3}+3 x^{2}-6 .\newlineThe absolute maximum value of f f over the closed interval [−2,5] [-2,5] occurs at what x x -value?\newlineChoose 11 answer:\newline(A) 55\newline(B) 00\newline(C) 22\newline(D) −2-2
  1. Find Derivative: To find the absolute maximum value of the function f(x)=−x3+3x2−6f(x)=-x^3+3x^2-6 over the interval [−2,5][-2,5], we need to find the critical points of the function within the interval and evaluate the function at the endpoints of the interval.\newlineFirst, we find the derivative of f(x)f(x) to locate the critical points.\newlinef′(x)=ddx(−x3+3x2−6)f'(x) = \frac{d}{dx} (-x^3 + 3x^2 - 6)\newlinef′(x)=−3x2+6xf'(x) = -3x^2 + 6x
  2. Locate Critical Points: Next, we set the derivative equal to zero to find the critical points.\newline−3x2+6x=0-3x^2 + 6x = 0\newlinex(−3x+6)=0x(-3x + 6) = 0\newlineThis gives us two solutions: x=0x = 0 and x=2x = 2.
  3. Evaluate Function: Now we need to evaluate the function f(x)f(x) at the critical points and at the endpoints of the interval.\newlineWe will calculate f(−2)f(-2), f(0)f(0), f(2)f(2), and f(5)f(5).\newlinef(−2)=−(−2)3+3(−2)2−6=−(−8)+3(4)−6=8+12−6=14f(-2) = -(-2)^3 + 3(-2)^2 - 6 = -(-8) + 3(4) - 6 = 8 + 12 - 6 = 14\newlinef(0)=−(0)3+3(0)2−6=−6f(0) = -(0)^3 + 3(0)^2 - 6 = -6\newlinef(2)=−(2)3+3(2)2−6=−8+3(4)−6=−8+12−6=−2f(2) = -(2)^3 + 3(2)^2 - 6 = -8 + 3(4) - 6 = -8 + 12 - 6 = -2\newlinef(5)=−(5)3+3(5)2−6=−125+3(25)−6=−125+75−6=−56f(5) = -(5)^3 + 3(5)^2 - 6 = -125 + 3(25) - 6 = -125 + 75 - 6 = -56
  4. Compare Values: Comparing the values of f(x)f(x) at the critical points and endpoints, we find that the largest value is f(−2)=14f(-2) = 14. Therefore, the absolute maximum value of f(x)f(x) over the interval [−2,5][-2,5] occurs at x=−2x = -2.

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