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Let
g(x)=2x^(3)-21x^(2)+60 x.
What is the absolute maximum value of
g over the closed interval
[0,6] ?
Choose 1 answer:
(A) 25
(B) 42
(C) 36
(D) 52

Let $g(x)=2 x^{3}-21 x^{2}+60 x$ . $\newline$ What is the absolute maximum value of $g$ over the closed interval $[0,6]$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $25$ $\newline$ (B) $42$ $\newline$ (C) $36$ $\newline$ (D) $52$

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Full solution

Q. Let

g(x)=2 x^{3}-21 x^{2}+60 x

.

\newline

What is the absolute maximum value of

g

over the closed interval

[0,6]

?

\newline

Choose

1

answer:

\newline

(A)

25

\newline

(B)

42

\newline

(C)

36

\newline

(D)

52

Find Critical Points: To find the absolute maximum, we need to find the critical points of $g(x)$ by taking the derivative and setting it to zero. $\newline$ $g'(x) = \frac{d}{dx} [2x^3 - 21x^2 + 60x]$
Calculate $g'(x)$ : Calculate $g'(x)$ : $g'(x) = 6x^2 - 42x + 60$
Set $g'(x)$ to zero: Set $g'(x)$ to zero and solve for x: $\newline$ $0 = 6x^2 - 42x + 60$
Factor the quadratic equation: Factor the quadratic equation: $\newline$ $0 = 6(x^2 - 7x + 10)$ $\newline$ $0 = 6(x - 5)(x - 2)$
Find critical points: Find the critical points: $x = 5$ and $x = 2$
Evaluate $g(x)$ : Evaluate $g(x)$ at the critical points and the endpoints of the interval $[0,6]$ :
$g(0) = 2(0)^3 - 21(0)^2 + 60(0) = 0$
$g(2) = 2(2)^3 - 21(2)^2 + 60(2) = 16 - 84 + 120 = 52$
$g(5) = 2(5)^3 - 21(5)^2 + 60(5) = 250 - 525 + 300 = 25$
$g(6) = 2(6)^3 - 21(6)^2 + 60(6) = 432 - 756 + 360 = 36$
Compare values: Compare the values to find the absolute maximum: $\newline$ g $(0) = 0$ , g $(2) = 52$ , g $(5) = 25$ , g $(6) = 36$ $\newline$ The absolute maximum value is $52$ at $x = 2$ .

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