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) 1 (C) 3 (D) 0

Find Derivative: To find the absolute minimum value of the function on the closed interval, we need to find the critical points of the function within the interval and evaluate the function at the endpoints of the interval. First, we find the derivative of the function 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 -12x + 12 = 0 x = 0 or x = 1Evaluate Function: We have two critical points within the interval: x = 0 and x = 1. Now we need to evaluate the function f(x) at these critical points and at the endpoints of the interval, x = -2 and x = 3. 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 = -59Compare Values: Now we compare the values of f(x) at x = -2, x = 0, x = 1, and x = 3 to find the smallest value, which will be the absolute minimum. f(-2) = 51 f(0) = -5 f(1) = -3 f(3) = -59 The smallest value is f(3) = -59.Find Absolute Minimum: The absolute minimum value of f(x) over the closed interval [-2, 3] occurs at x = 3.

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

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) 1
(C) 3
(D) 0

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) $1$ $\newline$ (C) $3$ $\newline$ (D) $0$

View step-by-step help

Home

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)

1

\newline

(C)

3

\newline

(D)

0

Find Derivative: To find the absolute minimum value of the function on the closed interval, we need to find the critical points of the function within the interval and evaluate the function at the endpoints of the interval. $\newline$ First, we find the derivative of the function $f(x) = -4x^3 + 6x^2 - 5$ . $\newline$ $f'(x) = \frac{d}{dx} (-4x^3 + 6x^2 - 5) = -12x^2 + 12x$ .
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 $-12x + 12 = 0$
$x = 0$ or $x = 1$
Evaluate Function: We have two critical points within the interval: $x = 0$ and $x = 1$ . Now we need to evaluate the function $f(x)$ at these critical points and at the endpoints of the interval, $x = -2$ and $x = 3$ .
$f(-2) = -4(-2)^3 + 6(-2)^2 - 5 = -4(-8) + 6(4) - 5 = 32 + 24 - 5 = 51$
$f(0) = -4(0)^3 + 6(0)^2 - 5 = -5$
$f(1) = -4(1)^3 + 6(1)^2 - 5 = -4 + 6 - 5 = -3$
$f(3) = -4(3)^3 + 6(3)^2 - 5 = -4(27) + 6(9) - 5 = -108 + 54 - 5 = -59$
Compare Values: Now we compare the values of $f(x)$ at $x = -2$ , $x = 0$ , $x = 1$ , and $x = 3$ to find the smallest value, which will be the absolute minimum. $\newline$ $f(-2) = 51$ $\newline$ $f(0) = -5$ $\newline$ $f(1) = -3$ $\newline$ $f(3) = -59$ $\newline$ The smallest value is $f(3) = -59$ .
Find Absolute Minimum: The absolute minimum value of $f(x)$ over the closed interval $[-2, 3]$ occurs at $x = 3$ .