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Let
h(x)=x^(3)+6x^(2)+2.
What is the absolute minimum value of
h over the closed interval
-6 <= x <= 2 ?
Choose 1 answer:
(A) 34
(B) 2
(C) -34
(D) -2

Let $h(x)=x^{3}+6 x^{2}+2$ . $\newline$ What is the absolute minimum value of $h$ over the closed interval $-6 \leq x \leq 2$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $34$ $\newline$ (B) $2$ $\newline$ (C) $-34$ $\newline$ (D) $-2$

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Full solution

Q. Let

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

.

\newline

What is the absolute minimum value of

h

over the closed interval

-6 \leq x \leq 2

?

\newline

Choose

1

answer:

\newline

(A)

34

\newline

(B)

2

\newline

(C)

-34

\newline

(D)

-2

Find Critical Points: To find the absolute minimum value of the function $h(x)$ on the closed interval $[-6, 2]$ , we first need to find the critical points of $h(x)$ within the interval. Critical points occur where the derivative $h'(x)$ is zero or undefined.
Calculate Derivative: Calculate the derivative of $h(x)$ to find $h'(x)$ : $h'(x) = \frac{d}{dx} (x^3 + 6x^2 + 2)$ $= 3x^2 + 12x$
Set Equal to Zero: Set the derivative equal to zero to find the critical points: $\newline$ $3x^2 + 12x = 0$ $\newline$ $x(3x + 12) = 0$ $\newline$ This gives us two solutions: $x = 0$ and $x = -4$ .
Check Interval: Check if the critical points are within the closed interval $[-6, 2]$ . Both $x = 0$ and $x = -4$ are within the interval, so we will evaluate $h(x)$ at these points as well as at the endpoints of the interval, $x = -6$ and $x = 2$ .
Evaluate at $-6$ : Evaluate $h(x)$ at $x = -6$ : $h(-6) = (-6)^3 + 6(-6)^2 + 2 = -216 + 216 + 2 = 2$
Evaluate at $-4$ : Evaluate $h(x)$ at $x = -4$ : $h(-4) = (-4)^3 + 6(-4)^2 + 2 = -64 + 96 + 2 = 34$
Evaluate at $0$ : Evaluate $h(x)$ at $x = 0$ : $h(0) = (0)^3 + 6(0)^2 + 2$ $= 0 + 0 + 2$ $= 2$
Evaluate at $2$ : Evaluate $h(x)$ at $x = 2$ : $h(2) = (2)^3 + 6(2)^2 + 2 = 8 + 24 + 2 = 34$
Compare Values: Compare the values of $h(x)$ at the critical points and the endpoints to determine the absolute minimum value: $h(-6) = 2$ $h(-4) = 34$ $h(0) = 2$ $h(2) = 34$ The absolute minimum value of $h(x)$ on the interval $[-6, 2]$ is $2$ , which occurs at both $x = -6$ and $x = 0$ .

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