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) -1 (B) 6 (C) 4 (D) 0

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. First, we find the derivative of h(x) to locate the critical points. h'(x) = d/dx (x^3 - 6x^2 + 8) h'(x) = 3x^2 - 12xLocate Critical Points: Next, we set the derivative equal to zero to find the critical points. 0 = 3x^2 - 12x 0 = 3x(x - 4) This gives us two critical points: x = 0 and x = 4.Evaluate h(x): Now we evaluate h(x) at the critical points and at the endpoints of the interval. h(-1) = (-1)^3 - 6(-1)^2 + 8 = -1 - 6 + 8 = 1 h(0) = (0)^3 - 6(0)^2 + 8 = 8 h(4) = (4)^3 - 6(4)^2 + 8 = 64 - 96 + 8 = -24 h(6) = (6)^3 - 6(6)^2 + 8 = 216 - 216 + 8 = 8Compare Values: We compare the values of h(x) at x = -1, x = 0, x = 4, and x = 6 to find the smallest value. h(-1) = 1 h(0) = 8 h(4) = -24 h(6) = 8 The smallest value is h(4) = -24, which means the absolute minimum 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) -1
(B) 6
(C) 4
(D) 0

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) $-1$ $\newline$ (B) $6$ $\newline$ (C) $4$ $\newline$ (D) $0$

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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)

-1

\newline

(B)

6

\newline

(C)

4

\newline

(D)

0

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. $\newline$ First, we find the derivative of $h(x)$ to locate the critical points. $\newline$ $h'(x) = \frac{d}{dx} (x^3 - 6x^2 + 8)$ $\newline$ $h'(x) = 3x^2 - 12x$
Locate Critical Points: Next, we set the derivative equal to zero to find the critical points. $\newline$ $0 = 3x^2 - 12x$ $\newline$ $0 = 3x(x - 4)$ $\newline$ This gives us two critical points: $x = 0$ and $x = 4$ .
Evaluate $h(x)$ : Now we evaluate $h(x)$ at the critical points and at the endpoints of the interval. $\newline$ $h(-1) = (-1)^3 - 6(-1)^2 + 8 = -1 - 6 + 8 = 1$ $\newline$ $h(0) = (0)^3 - 6(0)^2 + 8 = 8$ $\newline$ $h(4) = (4)^3 - 6(4)^2 + 8 = 64 - 96 + 8 = -24$ $\newline$ $h(6) = (6)^3 - 6(6)^2 + 8 = 216 - 216 + 8 = 8$
Compare Values: We compare the values of $h(x)$ at $x = -1$ , $x = 0$ , $x = 4$ , and $x = 6$ to find the smallest value. $\newline$ $h(-1) = 1$ $\newline$ $h(0) = 8$ $\newline$ $h(4) = -24$ $\newline$ $h(6) = 8$ $\newline$ The smallest value is $h(4) = -24$ , which means the absolute minimum occurs at $x = 4$ .