Let h(x)=2x^(3)+3x^(2)-12 x+5". " The absolute minimum value of h over the closed interval -3 <= x <= 2 occurs at what x value? Choose 1 answer: (A) 1 (B) 2 (C) -2 (D) -3

Calculate Derivative of h(x): To find the absolute minimum value of the function h(x) on the closed interval [-3, 2], 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.Find Critical Points: First, we calculate the derivative of h(x): h'(x) = d/dx [2x^3 + 3x^2 - 12x + 5] = 6x^2 + 6x - 12.Evaluate h(x) at Points: Next, we find the critical points by setting h'(x) to zero and solving for x: 6x^2 + 6x - 12 = 0. To simplify the equation, we can divide by 6: x^2 + x - 2 = 0.We factor the quadratic equation: (x + 2)(x - 1) = 0. This gives us two critical points: x = -2 and x = 1.Now we evaluate h(x) at the critical points and at the endpoints of the interval: h(-3), h(-2), h(1), and h(2).h(-3) = 2(-3)^3 + 3(-3)^2 - 12(-3) + 5 = -54 + 27 + 36 + 5 = 14.h(-2) = 2(-2)^3 + 3(-2)^2 - 12(-2) + 5 = -16 + 12 + 24 + 5 = 25.h(1) = 2(1)^3 + 3(1)^2 - 12(1) + 5 = 2 + 3 - 12 + 5 = -2.h(2) = 2(2)^3 + 3(2)^2 - 12(2) + 5 = 16 + 12 - 24 + 5 = 9.Comparing the values of h(x) at the critical points and endpoints, we find that the absolute minimum value occurs at x = 1, since h(1) = -2 is the smallest value.

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Let

h(x)=2x^(3)+3x^(2)-12 x+5". "
The absolute minimum value of
h over the closed interval
-3 <= x <= 2 occurs at what
x value?
Choose 1 answer:
(A) 1
(B) 2
(C) -2
(D) -3

Let $h(x)=2 x^{3}+3 x^{2}-12 x+5 \text {. }$ $\newline$ The absolute minimum value of $h$ over the closed interval $-3 \leq x \leq 2$ occurs at what $x$ value? $\newline$ Choose $1$ answer: $\newline$ (A) $1$ $\newline$ (B) $2$ $\newline$ (C) $-2$ $\newline$ (D) $-3$

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Algebra 2

Conjugate root theorems

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Q. Let

h(x)=2 x^{3}+3 x^{2}-12 x+5 \text {. }

\newline

The absolute minimum value of

h

over the closed interval

-3 \leq x \leq 2

occurs at what

x

value?

\newline

Choose

1

answer:

\newline

(A)

1

\newline

(B)

2

\newline

(C)

-2

\newline

(D)

-3

Calculate Derivative of $h(x)$ : To find the absolute minimum value of the function $h(x)$ on the closed interval $[-3, 2]$ , 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.
Find Critical Points: First, we calculate the derivative of $h(x)$ : $h'(x) = \frac{d}{dx} [2x^3 + 3x^2 - 12x + 5] = 6x^2 + 6x - 12.$
Evaluate $h(x)$ at Points: Next, we find the critical points by setting $h'(x)$ to zero and solving for $x$ : $6x^2 + 6x - 12 = 0$ . To simplify the equation, we can divide by $6$ : $x^2 + x - 2 = 0$ .
Evaluate $h(x)$ at Points: Next, we find the critical points by setting $h'(x)$ to zero and solving for $x$ : $6x^2 + 6x - 12 = 0$ . To simplify the equation, we can divide by $6$ : $x^2 + x - 2 = 0$ .We factor the quadratic equation: $(x + 2)(x - 1) = 0$ . This gives us two critical points: $x = -2$ and $x = 1$ .
Evaluate $h(x)$ at Points: Next, we find the critical points by setting $h'(x)$ to zero and solving for $x$ : $6x^2 + 6x - 12 = 0$ .To simplify the equation, we can divide by $6$ : $x^2 + x - 2 = 0$ .We factor the quadratic equation: $(x + 2)(x - 1) = 0$ .This gives us two critical points: $x = -2$ and $x = 1$ .Now we evaluate $h(x)$ at the critical points and at the endpoints of the interval: $h'(x)$ $0$ , $h'(x)$ $1$ , $h'(x)$ $2$ , and $h'(x)$ $3$ .
Evaluate $h(x)$ at Points: Next, we find the critical points by setting $h'(x)$ to zero and solving for $x$ : $6x^2 + 6x - 12 = 0$ .To simplify the equation, we can divide by $6$ : $x^2 + x - 2 = 0$ .We factor the quadratic equation: $(x + 2)(x - 1) = 0$ .This gives us two critical points: $x = -2$ and $x = 1$ .Now we evaluate $h(x)$ at the critical points and at the endpoints of the interval: $h'(x)$ $0$ , $h'(x)$ $1$ , $h'(x)$ $2$ , and $h'(x)$ $3$ . $h'(x)$ $4$ .
Evaluate $h(x)$ at Points: Next, we find the critical points by setting $h'(x)$ to zero and solving for $x$ : $6x^2 + 6x - 12 = 0$ .To simplify the equation, we can divide by $6$ : $x^2 + x - 2 = 0$ .We factor the quadratic equation: $(x + 2)(x - 1) = 0$ .This gives us two critical points: $x = -2$ and $x = 1$ .Now we evaluate $h(x)$ at the critical points and at the endpoints of the interval: $h'(x)$ $0$ , $h'(x)$ $1$ , $h'(x)$ $2$ , and $h'(x)$ $3$ . $h'(x)$ $4$ . $h'(x)$ $5$ .
Evaluate $h(x)$ at Points: Next, we find the critical points by setting $h'(x)$ to zero and solving for $x$ : $6x^2 + 6x - 12 = 0$ .To simplify the equation, we can divide by $6$ : $x^2 + x - 2 = 0$ .We factor the quadratic equation: $(x + 2)(x - 1) = 0$ .This gives us two critical points: $x = -2$ and $x = 1$ .Now we evaluate $h(x)$ at the critical points and at the endpoints of the interval: $h'(x)$ $0$ , $h'(x)$ $1$ , $h'(x)$ $2$ , and $h'(x)$ $3$ . $h'(x)$ $4$ . $h'(x)$ $5$ . $h'(x)$ $6$ .
Evaluate $h(x)$ at Points: Next, we find the critical points by setting $h'(x)$ to zero and solving for $x$ : $6x^2 + 6x - 12 = 0$ .To simplify the equation, we can divide by $6$ : $x^2 + x - 2 = 0$ .We factor the quadratic equation: $(x + 2)(x - 1) = 0$ .This gives us two critical points: $x = -2$ and $x = 1$ .Now we evaluate $h(x)$ at the critical points and at the endpoints of the interval: $h'(x)$ $0$ , $h'(x)$ $1$ , $h'(x)$ $2$ , and $h'(x)$ $3$ . $h'(x)$ $4$ . $h'(x)$ $5$ . $h'(x)$ $6$ . $h'(x)$ $7$ .
Evaluate $h(x)$ at Points: Next, we find the critical points by setting $h'(x)$ to zero and solving for $x$ : $6x^2 + 6x - 12 = 0$ .To simplify the equation, we can divide by $6$ : $x^2 + x - 2 = 0$ .We factor the quadratic equation: $(x + 2)(x - 1) = 0$ .This gives us two critical points: $x = -2$ and $x = 1$ .Now we evaluate $h(x)$ at the critical points and at the endpoints of the interval: $h'(x)$ $0$ , $h'(x)$ $1$ , $h'(x)$ $2$ , and $h'(x)$ $3$ . $h'(x)$ $4$ . $h'(x)$ $5$ . $h'(x)$ $6$ . $h'(x)$ $7$ .Comparing the values of $h(x)$ at the critical points and endpoints, we find that the absolute minimum value occurs at $x = 1$ , since $x$ $0$ is the smallest value.