Let h(x)=x^(3)-6x^(2)+8. The absolute minimum value of h over the closed interval -1 <= x <= 6 occurs at what x value? Choose 1 answer: (A) 0 (B) 4 (C) 6 (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. Let's find the derivative of h(x). h'(x) = d/dx [x^3 - 6x^2 + 8] = 3x^2 - 12xFind 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 = -1 - 6 + 8 = 1Evaluate h(x) at x = 0: Now let's evaluate h(x) at x = 0: h(0) = 0^3 - 6*0^2 + 8 = 0 - 0 + 8 = 8Evaluate h(x) at x = 4: Next, let's evaluate h(x) at x = 4: h(4) = 4^3 - 6*4^2 + 8 = 64 - 96 + 8 = -24Evaluate h(x) at x = 6: Finally, let's evaluate h(x) at x = 6: h(6) = 6^3 - 6*6^2 + 8 = 216 - 216 + 8 = 8Compare 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.

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

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 $x = 4$ .