Let $g(x)=-2 x^{3}+3 x^{2}+36 x$ . $\newline$ The absolute maximum value of $g$ over the closed interval $[-3,5]$ occurs at what $x$ -value? $\newline$ Choose $1$ answer: $\newline$ (A) $5$ $\newline$ (B) $3$ $\newline$ (C) $-3$ $\newline$ (D) $2$

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Euler's method

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Q. Let

g(x)=-2 x^{3}+3 x^{2}+36 x

\newline

The absolute maximum value of

g

over the closed interval

[-3,5]

occurs at what

x

-value?

\newline

Choose

1

answer:

\newline

(A)

5

\newline

(B)

3

\newline

(C)

-3

\newline

(D)

2

Find Derivative of $g(x)$ : To find the absolute maximum value of the function $g(x)$ on the closed interval $[-3,5]$ , we need to find the critical points of $g(x)$ within the interval and evaluate $g(x)$ at the endpoints of the interval.
Locate Critical Points: First, we find the derivative of $g(x)$ to locate the critical points. The derivative $g'(x)$ is given by: $\newline$ $g'(x) = \frac{d}{dx} (-2x^3 + 3x^2 + 36x) = -6x^2 + 6x + 36$ .
Evaluate Critical Points: Next, we find the critical points by setting $g'(x)$ to zero and solving for $x$ : $-6x^2 + 6x + 36 = 0$ .
Compare Values: We can factor out a $-6$ from the equation to simplify it: $-6(x^2 - x - 6) = 0$ .
Find Absolute Maximum: Now we factor the quadratic equation: $\newline$ $x^2 - x - 6 = (x - 3)(x + 2) = 0$ .
Find Absolute Maximum: Now we factor the quadratic equation: $x^2 - x - 6 = (x - 3)(x + 2) = 0$ . The solutions to the equation are $x = 3$ and $x = -2$ . These are the critical points where the function $g(x)$ could have a local maximum or minimum.
Find Absolute Maximum: Now we factor the quadratic equation: $\newline$ $x^2 - x - 6 = (x - 3)(x + 2) = 0$ .The solutions to the equation are $x = 3$ and $x = -2$ . These are the critical points where the function $g(x)$ could have a local maximum or minimum.We now evaluate $g(x)$ at the critical points and at the endpoints of the interval $[-3,5]$ : $\newline$ $g(-3) = -2(-3)^3 + 3(-3)^2 + 36(-3) = -54 + 27 - 108 = -135$ . $\newline$ $g(-2) = -2(-2)^3 + 3(-2)^2 + 36(-2) = 16 + 12 - 72 = -44$ . $\newline$ $g(3) = -2(3)^3 + 3(3)^2 + 36(3) = -54 + 27 + 108 = 81$ . $\newline$ $g(5) = -2(5)^3 + 3(5)^2 + 36(5) = -250 + 75 + 180 = 5$ .
Find Absolute Maximum: Now we factor the quadratic equation: $\newline$ $x^2 - x - 6 = (x - 3)(x + 2) = 0$ .The solutions to the equation are $x = 3$ and $x = -2$ . These are the critical points where the function $g(x)$ could have a local maximum or minimum.We now evaluate $g(x)$ at the critical points and at the endpoints of the interval $[-3,5]$ : $\newline$ $g(-3) = -2(-3)^3 + 3(-3)^2 + 36(-3) = -54 + 27 - 108 = -135$ . $\newline$ $g(-2) = -2(-2)^3 + 3(-2)^2 + 36(-2) = 16 + 12 - 72 = -44$ . $\newline$ $g(3) = -2(3)^3 + 3(3)^2 + 36(3) = -54 + 27 + 108 = 81$ . $\newline$ $g(5) = -2(5)^3 + 3(5)^2 + 36(5) = -250 + 75 + 180 = 5$ .Comparing the values of $g(x)$ at the critical points and endpoints, we find that the maximum value occurs at $x = 3$ , which is $x = 3$ $2$ .
Find Absolute Maximum: Now we factor the quadratic equation: $\newline$ $x^2 - x - 6 = (x - 3)(x + 2) = 0$ . The solutions to the equation are $x = 3$ and $x = -2$ . These are the critical points where the function $g(x)$ could have a local maximum or minimum. We now evaluate $g(x)$ at the critical points and at the endpoints of the interval $[-3,5]$ : $\newline$ $g(-3) = -2(-3)^3 + 3(-3)^2 + 36(-3) = -54 + 27 - 108 = -135$ . $\newline$ $g(-2) = -2(-2)^3 + 3(-2)^2 + 36(-2) = 16 + 12 - 72 = -44$ . $\newline$ $g(3) = -2(3)^3 + 3(3)^2 + 36(3) = -54 + 27 + 108 = 81$ . $\newline$ $g(5) = -2(5)^3 + 3(5)^2 + 36(5) = -250 + 75 + 180 = 5$ . Comparing the values of $g(x)$ at the critical points and endpoints, we find that the maximum value occurs at $x = 3$ , which is $81$ . Therefore, the absolute maximum value of $g(x)$ over the interval $[-3,5]$ occurs at $x = 3$ .