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) $2$ $\newline$ (C) $3$ $\newline$ (D) $-3$

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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)

2

\newline

(C)

3

\newline

(D)

-3

Calculate Derivative: 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. Critical points occur where the derivative $g'(x)$ is zero or undefined.
Find Critical Points: First, we calculate the derivative of $g(x)$ : $g'(x) = \frac{d}{dx} (-2x^3 + 3x^2 + 36x)$ $= -6x^2 + 6x + 36$
Evaluate Function: Next, we find the critical points by setting $g'(x)$ to zero and solving for $x$ :
$-6x^2 + 6x + 36 = 0$
Divide by $-6$ :
$x^2 - x - 6 = 0$
Factor the quadratic:
$(x - 3)(x + 2) = 0$
So, $x = 3$ or $x = -2$
Maximum Value: Now we need to evaluate $g(x)$ at the critical points and at the endpoints of the interval $[-3,5]$ : $\newline$ $g(-3)$ , $g(-2)$ , $g(3)$ , and $g(5)$ .
Maximum Value: Now we need to evaluate $g(x)$ at the critical points and at the endpoints of the interval $[-3,5]$ : $g(-3)$ , $g(-2)$ , $g(3)$ , and $g(5)$ .Evaluate $g(x)$ at $x = -3$ : $g(-3) = -2(-3)^3 + 3(-3)^2 + 36(-3) = -2(-27) + 3(9) - 108 = 54 + 27 - 108 = -27$
Maximum Value: Now we need to evaluate $g(x)$ at the critical points and at the endpoints of the interval $[-3,5]$ : $g(-3)$ , $g(-2)$ , $g(3)$ , and $g(5)$ .Evaluate $g(x)$ at $x = -3$ : $g(-3) = -2(-3)^3 + 3(-3)^2 + 36(-3) = -2(-27) + 3(9) - 108 = 54 + 27 - 108 = -27$ Evaluate $g(x)$ at $[-3,5]$ $0$ : $[-3,5]$ $1$
Maximum Value: Now we need to evaluate $g(x)$ at the critical points and at the endpoints of the interval $[-3,5]$ : $g(-3)$ , $g(-2)$ , $g(3)$ , and $g(5)$ .Evaluate $g(x)$ at $x = -3$ : $g(-3) = -2(-3)^3 + 3(-3)^2 + 36(-3) = -2(-27) + 3(9) - 108 = 54 + 27 - 108 = -27$ Evaluate $g(x)$ at $[-3,5]$ $0$ : $[-3,5]$ $1$ Evaluate $g(x)$ at $[-3,5]$ $3$ : $[-3,5]$ $4$
Maximum Value: Now we need to evaluate $g(x)$ at the critical points and at the endpoints of the interval $[-3,5]$ : $g(-3)$ , $g(-2)$ , $g(3)$ , and $g(5)$ .Evaluate $g(x)$ at $x = -3$ : $g(-3) = -2(-3)^3 + 3(-3)^2 + 36(-3) = -2(-27) + 3(9) - 108 = 54 + 27 - 108 = -27$ Evaluate $g(x)$ at $[-3,5]$ $0$ : $[-3,5]$ $1$ Evaluate $g(x)$ at $[-3,5]$ $3$ : $[-3,5]$ $4$ Evaluate $g(x)$ at $[-3,5]$ $6$ : $[-3,5]$ $7$
Maximum Value: Now we need to evaluate $g(x)$ at the critical points and at the endpoints of the interval $[-3,5]$ : $g(-3)$ , $g(-2)$ , $g(3)$ , and $g(5)$ .Evaluate $g(x)$ at $x = -3$ : $g(-3) = -2(-3)^3 + 3(-3)^2 + 36(-3) = -2(-27) + 3(9) - 108 = 54 + 27 - 108 = -27$ Evaluate $g(x)$ at $[-3,5]$ $0$ : $[-3,5]$ $1$ Evaluate $g(x)$ at $[-3,5]$ $3$ : $[-3,5]$ $4$ Evaluate $g(x)$ at $[-3,5]$ $6$ : $[-3,5]$ $7$ Comparing the values of $g(x)$ at $x = -3$ , $[-3,5]$ $0$ , $[-3,5]$ $3$ , and $[-3,5]$ $6$ , we find that the maximum value is $g(-3)$ $3$ , which occurs at $[-3,5]$ $3$ .