Let g(x)=x^(3)+1. The absolute minimum value of g over the closed interval -2 <= x <= 3 occurs at what x value? Choose 1 answer: (A) -2 (B) 0 (C) 3 (D) -7

Identify Critical Points: To find the absolute minimum value of the function g(x) = x^3 + 1 over the closed interval -2 <= x <= 3, we need to evaluate the function at the critical points and the endpoints of the interval.Evaluate Function at Critical Points: First, we find the critical points by taking the derivative of g(x) and setting it equal to zero. g'(x) = d/dx (x^3 + 1) = 3x^2. Setting g'(x) = 0 gives us 3x^2 = 0, which implies x = 0.Evaluate Function at Endpoints: Now we evaluate the function g(x) at the critical point x = 0 and at the endpoints of the interval, x = -2 and x = 3. g(-2) = (-2)^3 + 1 = -8 + 1 = -7. g(0) = (0)^3 + 1 = 0 + 1 = 1. g(3) = (3)^3 + 1 = 27 + 1 = 28.Compare Values and Determine Minimum: Comparing the values of g(x) at x = -2, x = 0, and x = 3, we see that the smallest value is g(-2) = -7. Therefore, the absolute minimum value of g(x) over the interval -2 <= x <= 3 occurs at x = -2.

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Let
g(x)=x^(3)+1.
The absolute minimum value of
g over the closed interval
-2 <= x <= 3 occurs at what
x value?
Choose 1 answer:
(A) -2
(B) 0
(C) 3
(D) -7

Let $g(x)=x^{3}+1$ . $\newline$ The absolute minimum value of $g$ over the closed interval $-2 \leq x \leq 3$ occurs at what $x$ value? $\newline$ Choose $1$ answer: $\newline$ (A) $-2$ $\newline$ (B) $0$ $\newline$ (C) $3$ $\newline$ (D) $-7$

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Math Problems

Algebra 2

Solve quadratic inequalities

Full solution

Q. Let

g(x)=x^{3}+1

\newline

The absolute minimum value of

g

over the closed interval

-2 \leq x \leq 3

occurs at what

x

value?

\newline

Choose

1

answer:

\newline

(A)

-2

\newline

(B)

0

\newline

(C)

3

\newline

(D)

-7

Identify Critical Points: To find the absolute minimum value of the function $g(x) = x^3 + 1$ over the closed interval $-2 \leq x \leq 3$ , we need to evaluate the function at the critical points and the endpoints of the interval.
Evaluate Function at Critical Points: First, we find the critical points by taking the derivative of $g(x)$ and setting it equal to zero. $g'(x) = \frac{d}{dx} (x^3 + 1) = 3x^2.$ Setting $g'(x) = 0$ gives us $3x^2 = 0$ , which implies $x = 0$ .
Evaluate Function at Endpoints: Now we evaluate the function $g(x)$ at the critical point $x = 0$ and at the endpoints of the interval, $x = -2$ and $x = 3$ .
$g(-2) = (-2)^3 + 1 = -8 + 1 = -7.$
$g(0) = (0)^3 + 1 = 0 + 1 = 1.$
$g(3) = (3)^3 + 1 = 27 + 1 = 28.$
Compare Values and Determine Minimum: Comparing the values of $g(x)$ at $x = -2$ , $x = 0$ , and $x = 3$ , we see that the smallest value is $g(-2) = -7$ . Therefore, the absolute minimum value of $g(x)$ over the interval $-2 \leq x \leq 3$ occurs at $x = -2$ .