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) $32$ $\newline$ (B) $\mathbf{1 8 0}$ $\newline$ (C) $\mathbf{1 0 8}$ $\newline$ (D) $0$

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Q. 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)

32

\newline

(B)

\mathbf{1 8 0}

\newline

(C)

\mathbf{1 0 8}

\newline

(D)

0

Find Derivative: 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.
Find Critical Points: First, we find the derivative of the function $f(x) = 2x^3 + 21x^2 + 36x$ , which will give us the slope of the tangent to the curve at any point $x$ . The derivative $f'(x)$ is given by: $\newline$ $f'(x) = \frac{d}{dx} (2x^3 + 21x^2 + 36x) = 6x^2 + 42x + 36$ .
Evaluate Function: Next, we find the critical points by setting the derivative equal to zero and solving for $x$ : $6x^2 + 42x + 36 = 0.$ To solve this quadratic equation, we can either factor it or use the quadratic formula. Let's try to factor it first.
Calculate Maximum Value: Factoring the quadratic equation, we get: $6(x^2 + 7x + 6) = 6(x + 1)(x + 6) = 0$ . This gives us two critical points: $x = -1$ and $x = -6$ .
Calculate Maximum Value: Factoring the quadratic equation, we get: $\newline$ $6(x^2 + 7x + 6) = 6(x + 1)(x + 6) = 0$ . $\newline$ This gives us two critical points: $x = -1$ and $x = -6$ .Now we evaluate the function $f(x)$ at the critical points and at the endpoints of the interval $[-8,0]$ : $\newline$ $f(-8)$ , $f(-6)$ , $f(-1)$ , and $f(0)$ .
Calculate Maximum Value: Factoring the quadratic equation, we get: $\newline$ $6(x^2 + 7x + 6) = 6(x + 1)(x + 6) = 0$ . $\newline$ This gives us two critical points: $x = -1$ and $x = -6$ .Now we evaluate the function $f(x)$ at the critical points and at the endpoints of the interval $[-8,0]$ : $\newline$ $f(-8)$ , $f(-6)$ , $f(-1)$ , and $f(0)$ .Evaluating $f(x)$ at $x = -1$ $0$ : $\newline$ $x = -1$ $1$ .
Calculate Maximum Value: Factoring the quadratic equation, we get: $\newline$ $6(x^2 + 7x + 6) = 6(x + 1)(x + 6) = 0$ . $\newline$ This gives us two critical points: $x = -1$ and $x = -6$ .Now we evaluate the function $f(x)$ at the critical points and at the endpoints of the interval $[-8,0]$ : $\newline$ $f(-8)$ , $f(-6)$ , $f(-1)$ , and $f(0)$ .Evaluating $f(x)$ at $x = -1$ $0$ : $\newline$ $x = -1$ $1$ .Evaluating $f(x)$ at $x = -6$ : $\newline$ $x = -1$ $4$ .
Calculate Maximum Value: Factoring the quadratic equation, we get: $\newline$ $6(x^2 + 7x + 6) = 6(x + 1)(x + 6) = 0$ . $\newline$ This gives us two critical points: $x = -1$ and $x = -6$ .Now we evaluate the function $f(x)$ at the critical points and at the endpoints of the interval $[-8,0]$ : $\newline$ $f(-8)$ , $f(-6)$ , $f(-1)$ , and $f(0)$ .Evaluating $f(x)$ at $x = -1$ $0$ : $\newline$ $x = -1$ $1$ .Evaluating $f(x)$ at $x = -6$ : $\newline$ $x = -1$ $4$ .Evaluating $f(x)$ at $x = -1$ : $\newline$ $x = -1$ $7$ .
Calculate Maximum Value: Factoring the quadratic equation, we get: $\newline$ $6(x^2 + 7x + 6) = 6(x + 1)(x + 6) = 0$ . $\newline$ This gives us two critical points: $x = -1$ and $x = -6$ .Now we evaluate the function $f(x)$ at the critical points and at the endpoints of the interval $[-8,0]$ : $\newline$ $f(-8)$ , $f(-6)$ , $f(-1)$ , and $f(0)$ .Evaluating $f(x)$ at $x = -1$ $0$ : $\newline$ $x = -1$ $1$ .Evaluating $f(x)$ at $x = -6$ : $\newline$ $x = -1$ $4$ .Evaluating $f(x)$ at $x = -1$ : $\newline$ $x = -1$ $7$ .Evaluating $f(x)$ at $x = -1$ $9$ : $\newline$ $x = -6$ $0$ .
Calculate Maximum Value: Factoring the quadratic equation, we get: $\newline$ $6(x^2 + 7x + 6) = 6(x + 1)(x + 6) = 0$ . $\newline$ This gives us two critical points: $x = -1$ and $x = -6$ .Now we evaluate the function $f(x)$ at the critical points and at the endpoints of the interval $[-8,0]$ : $\newline$ $f(-8)$ , $f(-6)$ , $f(-1)$ , and $f(0)$ .Evaluating $f(x)$ at $x = -1$ $0$ : $\newline$ $x = -1$ $1$ .Evaluating $f(x)$ at $x = -6$ : $\newline$ $x = -1$ $4$ .Evaluating $f(x)$ at $x = -1$ : $\newline$ $x = -1$ $7$ .Evaluating $f(x)$ at $x = -1$ $9$ : $\newline$ $x = -6$ $0$ .Comparing the values of $f(x)$ at $x = -1$ $0$ , $x = -6$ , $x = -1$ , and $x = -1$ $9$ , we find that the largest value is $x = -6$ $6$ .
Calculate Maximum Value: Factoring the quadratic equation, we get: $\newline$ $6(x^2 + 7x + 6) = 6(x + 1)(x + 6) = 0$ . $\newline$ This gives us two critical points: $x = -1$ and $x = -6$ .Now we evaluate the function $f(x)$ at the critical points and at the endpoints of the interval $[-8,0]$ : $\newline$ $f(-8)$ , $f(-6)$ , $f(-1)$ , and $f(0)$ .Evaluating $f(x)$ at $x = -1$ $0$ : $\newline$ $x = -1$ $1$ .Evaluating $f(x)$ at $x = -6$ : $\newline$ $x = -1$ $4$ .Evaluating $f(x)$ at $x = -1$ : $\newline$ $x = -1$ $7$ .Evaluating $f(x)$ at $x = -1$ $9$ : $\newline$ $x = -6$ $0$ .Comparing the values of $f(x)$ at $x = -1$ $0$ , $x = -6$ , $x = -1$ , and $x = -1$ $9$ , we find that the largest value is $x = -6$ $6$ .Therefore, the absolute maximum value of the function $f(x)$ over the closed interval $[-8,0]$ is $x = -6$ $9$ .