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Let
h(x)=-x^(3)+4.
What is the absolute maximum value of
h over the closed interval
[-2,2] ?
Choose 1 answer:
(A)
12
(B) 4
(C) -16
(D)
16

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) $4$ $\newline$ (C) $-16$ $\newline$ (D) $\mathbf{1 6}$

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

Full solution

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)

4

\newline

(C)

-16

\newline

(D)

\mathbf{1 6}

Find Critical Points: To find the absolute maximum value of $h(x)$ on the interval $[-2,2]$ , we need to evaluate the function at the critical points and the endpoints of the interval.
Evaluate Function: First, let's find the critical points by taking the derivative of $h(x)$ and setting it equal to zero. $\newline$ $h'(x) = \frac{d}{dx} (-x^3 + 4) = -3x^2$ $\newline$ Set $h'(x) = 0$ to find critical points: $\newline$ $-3x^2 = 0$ $\newline$ $x = 0$
Compare Values: Now we evaluate the function $h(x)$ at the critical point $x = 0$ and at the endpoints of the interval $x = -2$ and $x = 2$ .
$h(0) = -(0)^3 + 4 = 4$
$h(-2) = -(-2)^3 + 4 = -(-8) + 4 = 8 + 4 = 12$
$h(2) = -(2)^3 + 4 = -8 + 4 = -4$
Identify Absolute Maximum: Comparing the values of $h(x)$ at the critical point and the endpoints, we find that the absolute maximum value is $12$ , which occurs at $x = -2$ .
Identify Absolute Maximum: Comparing the values of $h(x)$ at the critical point and the endpoints, we find that the absolute maximum value is $12$ , which occurs at $x = -2$ .Therefore, the absolute maximum value of $h(x)$ over the interval $[-2,2]$ is $12$ .

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