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.
-$f ( x ) = \ln ( x - 1 ) , [ 2,4 ]$
Round to the nearest thousandth.
A) $2.820$
B) $3.885$
C) $3.820$
D) $\pm 2.820$

Question

Quizplus · Accepted Answer

The Mean Value Theorem states that there exists a $c$ in the interval $(a, b)$ such that $\frac{f(b) - f(a)}{b - a} = f'(c)$. For $f(x) = \ln(x-1)$, $f'(x) = \frac{1}{x-1}$. Calculating $\frac{f(4) - f(2)}{4 - 2}$ gives $\frac{\ln(3) - \ln(1)}{2} = \frac{\ln(3)}{2}$. Setting $\frac{1}{c-1} = \frac{\ln(3)}{2}$ and solving for $c$ gives $c = 2.820$.

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 $\frac { f ( b ) - f ( a ) } { b - a } = f ^ { \prime } ( c )$