Let g(x)=3x^(3)+8. What is the absolute minimum value of g over the closed interval -2 <= x <= 2 ? Choose 1 answer: (A) -16 (B) 16 (C) 8 (D) -8

Find Critical Points: To find the absolute minimum value of the function g(x) = 3x^3 + 8 on the interval [-2, 2], we first need to find the critical points of g(x) within the interval. Critical points occur where the derivative g'(x) is zero or undefined.Calculate Derivative: We calculate the derivative of g(x): g'(x) = d/dx (3x^3 + 8) = 9x^2.Set Equal to Zero: Set the derivative equal to zero to find critical points: 9x^2 = 0. Solving for x gives us x = 0 as the only critical point within the interval [-2, 2].Evaluate Function: Now we evaluate the function g(x) at the critical point and the endpoints of the interval to find the absolute minimum value. g(-2) = 3(-2)^3 + 8 = 3(-8) + 8 = -24 + 8 = -16. g(0) = 3(0)^3 + 8 = 0 + 8 = 8. g(2) = 3(2)^3 + 8 = 3(8) + 8 = 24 + 8 = 32.Compare Values: Comparing the values of g(x) at the critical point and the endpoints, we find that the smallest value is g(-2) = -16.Find Absolute Minimum: The absolute minimum value of the function g(x) = 3x^3 + 8 over the closed interval [-2, 2] is -16.

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Let
g(x)=3x^(3)+8.
What is the absolute minimum value of
g over the closed interval
-2 <= x <= 2 ?
Choose 1 answer:
(A) -16
(B) 16
(C) 8
(D) -8

Let $g(x)=3 x^{3}+8$ . $\newline$ What is the absolute minimum value of $g$ over the closed interval $-2 \leq x \leq 2$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $-16$ $\newline$ (B) $16$ $\newline$ (C) $8$ $\newline$ (D) $-8$

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Q. Let

g(x)=3 x^{3}+8

\newline

What is the absolute minimum value of

g

over the closed interval

-2 \leq x \leq 2

\newline

Choose

1

answer:

\newline

(A)

-16

\newline

(B)

16

\newline

(C)

8

\newline

(D)

-8

Find Critical Points: To find the absolute minimum value of the function $g(x) = 3x^3 + 8$ on the interval $[-2, 2]$ , we first need to find the critical points of $g(x)$ within the interval. Critical points occur where the derivative $g'(x)$ is zero or undefined.
Calculate Derivative: We calculate the derivative of $g(x)$ : $g'(x) = \frac{d}{dx} (3x^3 + 8) = 9x^2$ .
Set Equal to Zero: Set the derivative equal to zero to find critical points: $\newline$ $9x^2 = 0$ . $\newline$ Solving for $x$ gives us $x = 0$ as the only critical point within the interval $[-2, 2]$ .
Evaluate Function: Now we evaluate the function $g(x)$ at the critical point and the endpoints of the interval to find the absolute minimum value. $g(-2) = 3(-2)^3 + 8 = 3(-8) + 8 = -24 + 8 = -16.$ $g(0) = 3(0)^3 + 8 = 0 + 8 = 8.$ $g(2) = 3(2)^3 + 8 = 3(8) + 8 = 24 + 8 = 32.$
Compare Values: Comparing the values of $g(x)$ at the critical point and the endpoints, we find that the smallest value is $g(-2) = -16$ .
Find Absolute Minimum: The absolute minimum value of the function $g(x) = 3x^3 + 8$ over the closed interval $[-2, 2]$ is $-16$ .