Let $h(x)=2 x^{3}+3 x^{2}-12 x$ . $\newline$ What is the absolute minimum value of $h$ over the closed interval $-3 \leq x \leq 3$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $-32$ $\newline$ (B) $-45$ $\newline$ (C) $-7$ $\newline$ (D) $20$

View step-by-step help

Home

Math Problems

Calculus

Euler's method

Full solution

Q. Let

h(x)=2 x^{3}+3 x^{2}-12 x

\newline

What is the absolute minimum value of

h

over the closed interval

-3 \leq x \leq 3

\newline

Choose

1

answer:

\newline

(A)

-32

\newline

(B)

-45

\newline

(C)

-7

\newline

(D)

20

Calculate Derivative: To find the absolute minimum value of the function $h(x) = 2x^3 + 3x^2 - 12x$ over the closed interval $[-3, 3]$ , we first need to find the critical points of the function within the interval. Critical points occur where the derivative $h'(x)$ is zero or undefined. Let's calculate the derivative of $h(x)$ .
Find Critical Points: The derivative of $h(x)$ with respect to $x$ is $h'(x) = \frac{d}{dx} [2x^3 + 3x^2 - 12x]$ . Using the power rule, we get $h'(x) = 6x^2 + 6x - 12$ .
Evaluate Function: Next, we need to find the values of $x$ where $h'(x) = 0$ . Setting the derivative equal to zero gives us the equation $6x^2 + 6x - 12 = 0$ . We can simplify this by dividing the entire equation by $6$ , resulting in $x^2 + x - 2 = 0$ .
Compare Values: We can factor the quadratic equation $x^2 + x - 2 = 0$ to $(x + 2)(x - 1) = 0$ . This gives us two critical points: $x = -2$ and $x = 1$ .
Find Absolute Minimum: Now we need to evaluate the function $h(x)$ at the critical points and at the endpoints of the interval to find the absolute minimum. The endpoints are $x = -3$ and $x = 3$ . We will calculate $h(-3)$ , $h(-2)$ , $h(1)$ , and $h(3)$ .
Find Absolute Minimum: Now we need to evaluate the function $h(x)$ at the critical points and at the endpoints of the interval to find the absolute minimum. The endpoints are $x = -3$ and $x = 3$ . We will calculate $h(-3)$ , $h(-2)$ , $h(1)$ , and $h(3)$ .Evaluating $h(x)$ at $x = -3$ : $h(-3) = 2(-3)^3 + 3(-3)^2 - 12(-3) = 2(-27) + 3(9) + 36 = -54 + 27 + 36 = 9$ .
Find Absolute Minimum: Now we need to evaluate the function $h(x)$ at the critical points and at the endpoints of the interval to find the absolute minimum. The endpoints are $x = -3$ and $x = 3$ . We will calculate $h(-3)$ , $h(-2)$ , $h(1)$ , and $h(3)$ . Evaluating $h(x)$ at $x = -3$ : $h(-3) = 2(-3)^3 + 3(-3)^2 - 12(-3) = 2(-27) + 3(9) + 36 = -54 + 27 + 36 = 9$ . Evaluating $h(x)$ at $x = -3$ $1$ : $x = -3$ $2$ .
Find Absolute Minimum: Now we need to evaluate the function $h(x)$ at the critical points and at the endpoints of the interval to find the absolute minimum. The endpoints are $x = -3$ and $x = 3$ . We will calculate $h(-3)$ , $h(-2)$ , $h(1)$ , and $h(3)$ . Evaluating $h(x)$ at $x = -3$ : $h(-3) = 2(-3)^3 + 3(-3)^2 - 12(-3) = 2(-27) + 3(9) + 36 = -54 + 27 + 36 = 9$ . Evaluating $h(x)$ at $x = -3$ $1$ : $x = -3$ $2$ . Evaluating $h(x)$ at $x = -3$ $4$ : $x = -3$ $5$ .
Find Absolute Minimum: Now we need to evaluate the function $h(x)$ at the critical points and at the endpoints of the interval to find the absolute minimum. The endpoints are $x = -3$ and $x = 3$ . We will calculate $h(-3)$ , $h(-2)$ , $h(1)$ , and $h(3)$ . Evaluating $h(x)$ at $x = -3$ : $h(-3) = 2(-3)^3 + 3(-3)^2 - 12(-3) = 2(-27) + 3(9) + 36 = -54 + 27 + 36 = 9$ . Evaluating $h(x)$ at $x = -3$ $1$ : $x = -3$ $2$ . Evaluating $h(x)$ at $x = -3$ $4$ : $x = -3$ $5$ . Evaluating $h(x)$ at $x = 3$ : $x = -3$ $8$ .
Find Absolute Minimum: Now we need to evaluate the function $h(x)$ at the critical points and at the endpoints of the interval to find the absolute minimum. The endpoints are $x = -3$ and $x = 3$ . We will calculate $h(-3)$ , $h(-2)$ , $h(1)$ , and $h(3)$ . Evaluating $h(x)$ at $x = -3$ : $h(-3) = 2(-3)^3 + 3(-3)^2 - 12(-3) = 2(-27) + 3(9) + 36 = -54 + 27 + 36 = 9$ . Evaluating $h(x)$ at $x = -3$ $1$ : $x = -3$ $2$ . Evaluating $h(x)$ at $x = -3$ $4$ : $x = -3$ $5$ . Evaluating $h(x)$ at $x = 3$ : $x = -3$ $8$ . Comparing the values of $h(x)$ at the critical points and endpoints, we find that the smallest value is $x = 3$ $0$ . Therefore, the absolute minimum value of $h(x)$ over the interval $x = 3$ $2$ is $x = 3$ $3$ .