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) -7 (B) 3 (C) -2 (D) 0

Find Critical Points: To find the absolute minimum value of the function g(x) = x^3 + 1 over the closed interval -2 <= x <= 3, we first need to find the critical points of the function within the interval. Critical points occur where the derivative of the function is zero or undefined.Calculate Derivative: We calculate the derivative of g(x) with respect to x. g'(x) = d/dx (x^3 + 1) = 3x^2.Set Equal to Zero: Set the derivative equal to zero to find the critical points. 3x^2 = 0 x^2 = 0 x = 0Evaluate Function: The critical point within the interval -2 <= x <= 3 is x = 0. Now we need to evaluate the function g(x) at the critical point and at the endpoints of the interval to determine the absolute minimum value.Evaluate at -2: Evaluate g(x) at x = -2. g(-2) = (-2)^3 + 1 = -8 + 1 = -7.Evaluate at 0: Evaluate g(x) at x = 0. g(0) = 0^3 + 1 = 0 + 1 = 1.Evaluate at 3: Evaluate g(x) at x = 3. g(3) = 3^3 + 1 = 27 + 1 = 28.Compare Values: Compare the values of g(x) at x = -2, x = 0, and x = 3 to find the smallest value. g(-2) = -7, g(0) = 1, g(3) = 28. The smallest value is g(-2) = -7, which means the absolute minimum 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) -7
(B) 3
(C) -2
(D) 0

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) $-7$ $\newline$ (B) $3$ $\newline$ (C) $-2$ $\newline$ (D) $0$

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Math Problems

Algebra 2

Solve quadratic inequalities

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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)

-7

\newline

(B)

3

\newline

(C)

-2

\newline

(D)

0

Find 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 first need to find the critical points of the function within the interval. Critical points occur where the derivative of the function is zero or undefined.
Calculate Derivative: We calculate the derivative of $g(x)$ with respect to $x$ . $g'(x) = \frac{d}{dx} (x^3 + 1) = 3x^2.$
Set Equal to Zero: Set the derivative equal to zero to find the critical points. $\newline$ $3x^2 = 0$ $\newline$ $x^2 = 0$ $\newline$ $x = 0$
Evaluate Function: The critical point within the interval $-2 \leq x \leq 3$ is $x = 0$ . Now we need to evaluate the function $g(x)$ at the critical point and at the endpoints of the interval to determine the absolute minimum value.
Evaluate at $-2$ : Evaluate $g(x)$ at $x = -2$ . $g(-2) = (-2)^3 + 1 = -8 + 1 = -7$ .
Evaluate at $0$ : Evaluate $g(x)$ at $x = 0$ . $\newline$ $g(0) = 0^3 + 1 = 0 + 1 = 1$ .
Evaluate at $3$ : Evaluate $g(x)$ at $x = 3$ . $g(3) = 3^3 + 1 = 27 + 1 = 28$ .
Compare Values: Compare the values of $g(x)$ at $x = -2$ , $x = 0$ , and $x = 3$ to find the smallest value. $g(-2) = -7$ , $g(0) = 1$ , $g(3) = 28$ . The smallest value is $g(-2) = -7$ , which means the absolute minimum occurs at $x = -2$ .