Let $h(x)=x^{3}-6 x^{2}+8$ . $\newline$ The absolute minimum value of $h$ over the closed interval $-1 \leq x \leq 6$ occurs at what $x$ value? $\newline$ Choose $1$ answer: $\newline$ (A) $0$ $\newline$ (B) $4$ $\newline$ (C) $6$ $\newline$ (D) $-1$

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Q. Let

h(x)=x^{3}-6 x^{2}+8

\newline

The absolute minimum value of

h

over the closed interval

-1 \leq x \leq 6

occurs at what

x

value?

\newline

Choose

1

answer:

\newline

(A)

0

\newline

(B)

4

\newline

(C)

6

\newline

(D)

-1

Find Derivative of h(x): To find the absolute minimum value of the function $h(x)$ over the closed interval $[-1, 6]$ , we need to find the critical points of $h(x)$ within the interval and evaluate $h(x)$ at the endpoints of the interval. Critical points occur where the derivative $h'(x)$ is zero or undefined. $\newline$ Let's find the derivative of $h(x)$ . $\newline$ $h'(x) = \frac{d}{dx} [x^3 - 6x^2 + 8]$ $\newline$ = $3x^2 - 12x$
Find Critical Points: Now we need to find the values of $x$ where $h'(x) = 0$ . Setting $h'(x)$ to zero gives us: $3x^2 - 12x = 0$ $x(3x - 12) = 0$ This gives us two solutions: $x = 0$ and $x = 4$ .
Evaluate $h(x)$ at Critical Points and Endpoints: We need to check if these critical points are within the interval $[-1, 6]$ . Both $x = 0$ and $x = 4$ are within the interval, so we will evaluate $h(x)$ at $x = 0$ , $x = 4$ , and at the endpoints $x = -1$ and $x = 6$ .
Evaluate $h(x)$ at $x = -1$ : Let's evaluate $h(x)$ at $x = -1$ :
$h(-1) = (-1)^3 - 6(-1)^2 + 8$
$\quad\; = -1 - 6 + 8$
$\quad\; = 1$
Evaluate $h(x)$ at $x = 0$ : Now let's evaluate $h(x)$ at $x = 0$ :
$h(0) = 0^3 - 6\cdot0^2 + 8$
$\phantom{h(0)} = 0 - 0 + 8$
$\phantom{h(0)} = 8$
Evaluate $h(x)$ at $x = 4$ : Next, let's evaluate $h(x)$ at $x = 4$ :
$h(4) = 4^3 - 6\times4^2 + 8$
$\phantom{h(4)} = 64 - 96 + 8$
$\phantom{h(4)} = -24$
Evaluate $h(x)$ at $x = 6$ : Finally, let's evaluate $h(x)$ at $x = 6$ :
$h(6) = 6^3 - 6\times6^2 + 8$
$\phantom{h(6)} = 216 - 216 + 8$
$\phantom{h(6)} = 8$
Compare Values and Determine Absolute Minimum: Comparing the values of $h(x)$ at $x = -1$ , $x = 0$ , $x = 4$ , and $x = 6$ , we find that the smallest value is $h(4) = -24$ . Therefore, the absolute minimum value of $h(x)$ over the interval $[-1, 6]$ occurs at $x = 4$ .