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Let f(x)=x^(3)-3x^(2)+12. What is the absolute maximum value of f over the closed interval [-2,4] ? Choose 1 answer: (A) 38 (B) 28 (C) -12 (D) 12

Let f(x)=x3−3x2+12 f(x)=x^{3}-3 x^{2}+12 .\newlineWhat is the absolute maximum value of f f over the closed interval [−2,4] [-2,4] ?\newlineChoose 11 answer:\newline(A) 3838\newline(B) 2828\newline(C) −12-12\newline(D) 12 \mathbf{1 2}

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Full solution

Q. Let f(x)=x3−3x2+12 f(x)=x^{3}-3 x^{2}+12 .\newlineWhat is the absolute maximum value of f f over the closed interval [−2,4] [-2,4] ?\newlineChoose 11 answer:\newline(A) 3838\newline(B) 2828\newline(C) −12-12\newline(D) 12 \mathbf{1 2}
  1. Find Derivative: To find the absolute maximum value of the function on the closed interval, we need to evaluate the function at critical points and endpoints of the interval. Critical points occur where the derivative is zero or undefined. First, we find the derivative of f(x)f(x).f′(x)=ddx(x3−3x2+12)=3x2−6xf'(x) = \frac{d}{dx} (x^3 - 3x^2 + 12) = 3x^2 - 6x.
  2. Find Critical Points: Next, we find the critical points by setting the derivative equal to zero and solving for xx.0=3x2−6x0 = 3x^2 - 6x0=x(3x−6)0 = x(3x - 6)x=0x = 0 or x=2x = 2.
  3. Evaluate Function: Now we evaluate the function f(x)f(x) at the critical points and the endpoints of the interval [−2,4][-2, 4].
    f(−2)=(−2)3−3(−2)2+12=−8−12+12=−8f(-2) = (-2)^3 - 3(-2)^2 + 12 = -8 - 12 + 12 = -8
    f(0)=(0)3−3(0)2+12=12f(0) = (0)^3 - 3(0)^2 + 12 = 12
    f(2)=(2)3−3(2)2+12=8−12+12=8f(2) = (2)^3 - 3(2)^2 + 12 = 8 - 12 + 12 = 8
    f(4)=(4)3−3(4)2+12=64−48+12=28f(4) = (4)^3 - 3(4)^2 + 12 = 64 - 48 + 12 = 28
  4. Compare Values: We compare the values of f(x)f(x) at the critical points and endpoints to find the absolute maximum. The values are −8-8, 1212, 88, and 2828. The largest value among these is 2828.

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