Find the value or values of $c$ that satisfy the equation $\frac { f ( b ) - f ( a ) } { b - a } = f ^ { \prime } ( c )$ in the conclusion of the Mean Value Theorem for the function and interval.
-$$f ( x ) = x + \frac { 18 } { x } , [ 2,9 ]$$
A) 2,9
B) $3 \sqrt { 2 }$
C) $0,3 \sqrt { 2 }$
D) $- 3 \sqrt { 2 } , 3 \sqrt { 2 }$

Question

Quizplus · Accepted Answer

By the Mean Value Theorem, there exists a value c in the interval [2, 9] such that $\frac{f(9) - f(2)}{9-2} = f'(c)$. We have:

\begin{align*}
\frac{f(9) - f(2)}{9-2} &= \frac{\left(9+\frac{18}{9}\right) - \left(2+\frac{18}{2}\right)}{7} \
&= \frac{11}{21}.
\end{align*}

To find $f'(x)$, we differentiate the function:

$$f'(x) = 1 - \frac{18}{x^2}$$

Setting this equal to $\frac{11}{21}$, we get:

\begin{align*}
1 - \frac{18}{x^2} &= \frac{11}{21} \
\frac{18}{x^2} &= \frac{10}{21} \
x^2 &= \frac{37}{3} \
x &= \pm 3\sqrt{2}.
\end{align*}

Since $c$ must be in the interval [2, 9], the only value that works is $c = 3\sqrt{2}$. Therefore, the answer is B.

Find the Value or Values Of ccc That Satisfy the Equation f(b)−f(a)b−a=f′(c)\frac { f ( b ) - f ( a ) } { b - a } = f ^ { \prime } ( c )b−af(b)−f(a)​=f′(c)

Find the Value or Values Of $c$ That Satisfy the Equation $\frac { f ( b ) - f ( a ) } { b - a } = f ^ { \prime } ( c )$