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Let 
g(x)=x^(3)-12 x+7.
The absolute maximum value of 
g over the closed interval 
[-4,5] occurs at what 
x-value?
Choose 1 answer:
(A) 5
(B) -2
(C) 2
(D) -4

Let g(x)=x312x+7 g(x)=x^{3}-12 x+7 .\newlineThe absolute maximum value of g g over the closed interval [4,5] [-4,5] occurs at what x x -value?\newlineChoose 11 answer:\newline(A) 55\newline(B) 2-2\newline(C) 22\newline(D) 4-4

Full solution

Q. Let g(x)=x312x+7 g(x)=x^{3}-12 x+7 .\newlineThe absolute maximum value of g g over the closed interval [4,5] [-4,5] occurs at what x x -value?\newlineChoose 11 answer:\newline(A) 55\newline(B) 2-2\newline(C) 22\newline(D) 4-4
  1. Calculate Derivative: To find the absolute maximum value of the function g(x)=x312x+7g(x) = x^3 - 12x + 7 on the closed interval [4,5][-4, 5], we first need to find the critical points of g(x)g(x) within the interval. Critical points occur where the derivative g(x)g'(x) is zero or undefined.\newlineLet's calculate the derivative g(x)g'(x).\newlineg(x)=ddx(x312x+7)=3x212.g'(x) = \frac{d}{dx} (x^3 - 12x + 7) = 3x^2 - 12.
  2. Find Critical Points: Now we set the derivative equal to zero to find the critical points.\newline0=3x2120 = 3x^2 - 12\newlinex2=4x^2 = 4\newlinex=±2x = \pm2\newlineBoth x=2x = 2 and x=2x = -2 are within the interval [4,5][-4, 5].
  3. Evaluate Function: Next, we evaluate the function g(x)g(x) at the critical points and at the endpoints of the interval to determine the absolute maximum.g(4)=(4)312(4)+7=64+48+7=9g(-4) = (-4)^3 - 12(-4) + 7 = -64 + 48 + 7 = -9g(2)=(2)312(2)+7=8+24+7=23g(-2) = (-2)^3 - 12(-2) + 7 = -8 + 24 + 7 = 23g(2)=(2)312(2)+7=824+7=9g(2) = (2)^3 - 12(2) + 7 = 8 - 24 + 7 = -9g(5)=(5)312(5)+7=12560+7=72g(5) = (5)^3 - 12(5) + 7 = 125 - 60 + 7 = 72
  4. Determine Absolute Maximum: Comparing the values of g(x)g(x) at the critical points and endpoints, we find that the absolute maximum value occurs at x=5x = 5, which is 7272.

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