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) 8 (C) -16 (D) -8

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

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

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) 8
(C) -16
(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) $8$ $\newline$ (C) $-16$ $\newline$ (D) $-8$

View step-by-step help

Home

Math Problems

Algebra 2

Conjugate root theorems

Full solution

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)

8

\newline

(C)

-16

\newline

(D)

-8

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