Let $h(x)=-x^{3}+4$ . $\newline$ What is the absolute maximum value of $h$ over the closed interval $[-2,2]$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $\mathbf{1 2}$ $\newline$ (B) $\mathbf{1 6}$ $\newline$ (C) $4$ $\newline$ (D) $-16$

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Algebra 2

Evaluate expression when a complex numbers and a variable term is given

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Q. Let

h(x)=-x^{3}+4

\newline

What is the absolute maximum value of

h

over the closed interval

[-2,2]

\newline

Choose

1

answer:

\newline

(A)

\mathbf{1 2}

\newline

(B)

\mathbf{1 6}

\newline

(C)

4

\newline

(D)

-16

Find Critical Points: To find the absolute maximum value of the function $h(x) = -x^3 + 4$ over the closed interval $[-2,2]$ , we need to evaluate the function at the critical points and the endpoints of the interval.
Derivative Calculation: First, let's find the critical points by taking the derivative of $h(x)$ and setting it equal to zero. $h'(x) = \frac{d}{dx} (-x^3 + 4) = -3x^2.$
Evaluate Critical Points: Now, we set the derivative equal to zero to find the critical points: $\newline$ $-3x^2 = 0$ $\newline$ $x^2 = 0$ $\newline$ $x = 0$ .
Evaluate Endpoints: The only critical point within the interval $[-2,2]$ is $x = 0$ . Now we will evaluate the function $h(x)$ at the critical point and at the endpoints of the interval.
Compare Values: Evaluating $h(x)$ at $x = 0$ : $h(0) = -(0)^3 + 4 = 4$ .
Compare Values: Evaluating $h(x)$ at $x = 0$ :
$h(0) = -(0)^3 + 4 = 4$ .Evaluating $h(x)$ at the left endpoint $x = -2$ :
$h(-2) = -(-2)^3 + 4 = -(-8) + 4 = 8 + 4 = 12$ .
Compare Values: Evaluating $h(x)$ at $x = 0$ :
$h(0) = -(0)^3 + 4 = 4$ .Evaluating $h(x)$ at the left endpoint $x = -2$ :
$h(-2) = -(-2)^3 + 4 = -(-8) + 4 = 8 + 4 = 12$ .Evaluating $h(x)$ at the right endpoint $x = 2$ :
$h(2) = -(2)^3 + 4 = -8 + 4 = -4$ .
Compare Values: Evaluating $h(x)$ at $x = 0$ :
$h(0) = -(0)^3 + 4 = 4$ .Evaluating $h(x)$ at the left endpoint $x = -2$ :
$h(-2) = -(-2)^3 + 4 = -(-8) + 4 = 8 + 4 = 12$ .Evaluating $h(x)$ at the right endpoint $x = 2$ :
$h(2) = -(2)^3 + 4 = -8 + 4 = -4$ .Comparing the values of $h(x)$ at the critical point and the endpoints, we find that the absolute maximum value is $x = 0$ $0$ , which occurs at $x = -2$ .