Let f(x)=2x^(3)+21x^(2)+36 x. What is the absolute maximum value of f over the closed interval [-8,0] ? Choose 1 answer: (A) 0 (B) 108 (C) 180 (D) 32

Question

Let  f(x)=2x^(3)+21x^(2)+36 x. What is the absolute maximum value of  f over the closed interval  [-8,0] ? Choose 1 answer: (A) 0 (B) 108 (C) 180 (D) 32

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Find Derivative and Critical Points: To find the absolute maximum value of the function on the closed interval [-8,0], we need to find the critical points of the function within the interval and evaluate the function at these points and at the endpoints of the interval.Factor Quadratic Equation: First, we find the derivative of the function f(x) = 2x^3 + 21x^2 + 36x, which will give us the critical points where the slope of the tangent is zero or where the derivative does not exist. f'(x) = d/dx (2x^3 + 21x^2 + 36x) = 6x^2 + 42x + 36Evaluate Function at Critical Points: Next, we find the critical points by setting the derivative equal to zero and solving for x. 0 = 6x^2 + 42x + 36 To solve this quadratic equation, we can factor it or use the quadratic formula. Let's try to factor it first. 0 = 6(x^2 + 7x + 6) 0 = 6(x + 1)(x + 6) The solutions are x = -1 and x = -6.Compare Values: Now we evaluate the function f(x) at the critical points and at the endpoints of the interval. f(-8), f(-6), f(-1), and f(0) need to be calculated. f(-8) = 2(-8)^3 + 21(-8)^2 + 36(-8) = -1024 + 1344 - 288 = 32 f(-6) = 2(-6)^3 + 21(-6)^2 + 36(-6) = -432 + 756 - 216 = 108 f(-1) = 2(-1)^3 + 21(-1)^2 + 36(-1) = -2 + 21 - 36 = -17 f(0) = 2(0)^3 + 21(0)^2 + 36(0) = 0Find Absolute Maximum Value: Comparing the values we calculated, we find that the largest value is f(-6) = 108.The absolute maximum value of the function f(x) over the closed interval [-8,0] is 108.

Let $f(x)=2 x^{3}+21 x^{2}+36 x$ . $\newline$ What is the absolute maximum value of $f$ over the closed interval $[-8,0]$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $0$ $\newline$ (B) $108$ $\newline$ (C) $180$ $\newline$ (D) $32$

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Let f(x)=2x3+21x2+36x f(x)=2 x^{3}+21 x^{2}+36 x f(x)=2x3+21x2+36x.\newlineWhat is the absolute maximum value of f f f over the closed interval [−8,0] [-8,0] [−8,0] ?\newlineChoose 111 answer:\newline(A) 000\newline(B) 108108108\newline(C) 180180180\newline(D) 323232

Let $f(x)=2 x^{3}+21 x^{2}+36 x$ . $\newline$ What is the absolute maximum value of $f$ over the closed interval $[-8,0]$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $0$ $\newline$ (B) $108$ $\newline$ (C) $180$ $\newline$ (D) $32$