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Let
g(x)=x^(3)+12x^(2)+36 x and let
c be the number that satisfies the Mean Value Theorem for
g on the interval
-8 <= x <= -2.
What is
c ?
Choose 1 answer:
(A) -7
(B) -6
(c) -3
(b) -1

Let $g(x)=x^{3}+12x^{2}+36x$ and let $c$ be the number that satisfies the Mean Value Theorem for $g$ on the interval $-8 \leq x \leq -2$ . $\newline$ What is $c$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $-7$ $\newline$ (B) $-6$ $\newline$ (C) $-3$ $\newline$ (D) $-1$

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Full solution

Q. Let

g(x)=x^{3}+12x^{2}+36x

and let

c

be the number that satisfies the Mean Value Theorem for

g

on the interval

-8 \leq x \leq -2

.

\newline

What is

c

?

\newline

Choose

1

answer:

\newline

(A)

-7

\newline

(B)

-6

\newline

(C)

-3

\newline

(D)

-1

Calculate Derivative of $g(x)$ : Calculate the derivative of $g(x)$ to apply the Mean Value Theorem. $\newline$ $g(x) = x^3 + 12x^2 + 36x$ $\newline$ $g'(x) = 3x^2 + 24x + 36$
Apply Mean Value Theorem: Apply the Mean Value Theorem, which states there exists a $c$ in $(-8, -2)$ such that $g'(-8) = g'(-2)$ . $\newline$ Calculate $g'(-8)$ and $g'(-2)$ : $\newline$ $g'(-8) = 3(-8)^2 + 24(-8) + 36 = 192 - 192 + 36 = 36$ $\newline$ $g'(-2) = 3(-2)^2 + 24(-2) + 36 = 12 - 48 + 36 = 0$
Find $c$ for $g'(-8) = g'(-2)$ : Find $c$ such that $g'(-8) = g'(-2)$ . $\newline$ Since $g'(-8) \neq g'(-2)$ , there's a mistake in the calculation of $g'(-2)$ . $\newline$ Recalculate $g'(-2)$ : $\newline$ $g'(-2) = 3(-2)^2 + 24(-2) + 36 = 12 - 48 + 36 = 0$

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