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Let
f(x)=x^(3)-3x^(2)+12.
What is the absolute maximum value of
f over the closed interval
[-2,4] ?
Choose 1 answer:
(A) 38
(B)
12
(C) -12
(D) 28

Let $f(x)=x^{3}-3 x^{2}+12$ . $\newline$ What is the absolute maximum value of $f$ over the closed interval $[-2,4]$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $38$ $\newline$ (B) $\mathbf{1 2}$ $\newline$ (C) $-12$ $\newline$ (D) $28$

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Conjugate root theorems

Full solution

Q. Let

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

.

\newline

What is the absolute maximum value of

f

over the closed interval

[-2,4]

?

\newline

Choose

1

answer:

\newline

(A)

38

\newline

(B)

\mathbf{1 2}

\newline

(C)

-12

\newline

(D)

28

Find Derivative: To find the absolute maximum value of the function on the closed interval $[-2, 4]$ , we need to evaluate the function at the critical points within the interval and at the endpoints of the interval. The critical points are where the first derivative is zero or undefined. $\newline$ First, we find the first derivative of $f(x)$ : $\newline$ $f'(x) = \frac{d}{dx} (x^3 - 3x^2 + 12)$ $\newline$ $= 3x^2 - 6x$
Find Critical Points: Next, we set the first derivative equal to zero to find the critical points: $\newline$ $3x^2 - 6x = 0$ $\newline$ $x(3x - 6) = 0$ $\newline$ $x = 0$ or $x = 2$ $\newline$ These are the critical points within the interval $[-2, 4]$ .
Evaluate Function: Now we evaluate the function $f(x)$ at the critical points and at the endpoints of the interval: $f(-2) = (-2)^3 - 3(-2)^2 + 12 = -8 - 12 + 12 = -8$ $f(0) = (0)^3 - 3(0)^2 + 12 = 12$ $f(2) = (2)^3 - 3(2)^2 + 12 = 8 - 12 + 12 = 8$ $f(4) = (4)^3 - 3(4)^2 + 12 = 64 - 48 + 12 = 28$
Compare Values: We compare the values of $f(x)$ at these points to determine the absolute maximum: $\newline$ $f(-2) = -8$ $\newline$ $f(0) = 12$ $\newline$ $f(2) = 8$ $\newline$ $f(4) = 28$ $\newline$ The largest value is $f(4) = 28$ , which is the absolute maximum value on the interval $[-2, 4]$ .

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