Let f(x)=-4x^(3)+6x^(2)-5. The absolute minimum value of f over the closed interval -2 <= x <= 3 occurs at what x value? Choose 1 answer: (A) -2 (B) 3 (C) 0 (D) 1

Find Critical Points: To find the absolute minimum value of the function on the closed interval [-2, 3], we first need to find the critical points of the function within the interval. Critical points occur where the derivative of the function is zero or undefined.Calculate Derivative: We calculate the derivative of f(x) = -4x^3 + 6x^2 - 5. f'(x) = d/dx (-4x^3 + 6x^2 - 5) = -12x^2 + 12x.Find Critical Points: Next, we find the critical points by setting the derivative equal to zero and solving for x. -12x^2 + 12x = 0 x(-12x + 12) = 0 x = 0 or x = 1.Evaluate Function: We now have two critical points, x = 0 and x = 1, within the interval [-2, 3]. We also need to evaluate the function at the endpoints of the interval, which are x = -2 and x = 3.Compare Values: We evaluate the function at the critical points and the endpoints: f(-2) = -4(-2)^3 + 6(-2)^2 - 5 = -4(-8) + 6(4) - 5 = 32 + 24 - 5 = 51. f(0) = -4(0)^3 + 6(0)^2 - 5 = -5. f(1) = -4(1)^3 + 6(1)^2 - 5 = -4 + 6 - 5 = -3. f(3) = -4(3)^3 + 6(3)^2 - 5 = -4(27) + 6(9) - 5 = -108 + 54 - 5 = -59.Absolute Minimum Value: Comparing the values of f(x) at the critical points and endpoints, we find that the absolute minimum value occurs at x = 1, since f(1) = -3 is the smallest value.The answer to the question prompt is that the absolute minimum value of the function f(x) = -4x^3 + 6x^2 - 5 over the interval [-2, 3] occurs at x = 1.

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Let
f(x)=-4x^(3)+6x^(2)-5.
The absolute minimum value of
f over the closed interval
-2 <= x <= 3 occurs at what
x value?
Choose 1 answer:
(A) -2
(B) 3
(C) 0
(D) 1

Let $f(x)=-4 x^{3}+6 x^{2}-5$ . $\newline$ The absolute minimum value of $f$ over the closed interval $-2 \leq x \leq 3$ occurs at what $x$ value? $\newline$ Choose $1$ answer: $\newline$ (A) $-2$ $\newline$ (B) $3$ $\newline$ (C) $0$ $\newline$ (D) $1$

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Math Problems

Algebra 2

Solve quadratic inequalities

Full solution

Q. Let

f(x)=-4 x^{3}+6 x^{2}-5

\newline

The absolute minimum value of

f

over the closed interval

-2 \leq x \leq 3

occurs at what

x

value?

\newline

Choose

1

answer:

\newline

(A)

-2

\newline

(B)

3

\newline

(C)

0

\newline

(D)

1

Find Critical Points: To find the absolute minimum value of the function on the closed interval $[-2, 3]$ , we first need to find the critical points of the function within the interval. Critical points occur where the derivative of the function is zero or undefined.
Calculate Derivative: We calculate the derivative of $f(x) = -4x^3 + 6x^2 - 5$ . $f'(x) = \frac{d}{dx} (-4x^3 + 6x^2 - 5) = -12x^2 + 12x$ .
Find Critical Points: Next, we find the critical points by setting the derivative equal to zero and solving for $x$ . $-12x^2 + 12x = 0$ $x(-12x + 12) = 0$ $x = 0$ or $x = 1$ .
Evaluate Function: We now have two critical points, $x = 0$ and $x = 1$ , within the interval $[-2, 3]$ . We also need to evaluate the function at the endpoints of the interval, which are $x = -2$ and $x = 3$ .
Compare Values: We evaluate the function at the critical points and the endpoints: $\newline$ $f(-2) = -4(-2)^3 + 6(-2)^2 - 5 = -4(-8) + 6(4) - 5 = 32 + 24 - 5 = 51$ . $\newline$ $f(0) = -4(0)^3 + 6(0)^2 - 5 = -5$ . $\newline$ $f(1) = -4(1)^3 + 6(1)^2 - 5 = -4 + 6 - 5 = -3$ . $\newline$ $f(3) = -4(3)^3 + 6(3)^2 - 5 = -4(27) + 6(9) - 5 = -108 + 54 - 5 = -59$ .
Absolute Minimum Value: Comparing the values of $f(x)$ at the critical points and endpoints, we find that the absolute minimum value occurs at $x = 1$ , since $f(1) = -3$ is the smallest value.
Absolute Minimum Value: Comparing the values of $f(x)$ at the critical points and endpoints, we find that the absolute minimum value occurs at $x = 1$ , since $f(1) = -3$ is the smallest value.The answer to the question prompt is that the absolute minimum value of the function $f(x) = -4x^3 + 6x^2 - 5$ over the interval $[-2, 3]$ occurs at $x = 1$ .