The Mean Value Theorem (MVT) states that if a function $f(x)$ is continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$, then there exists a number $c$ in $(a, b)$ such that:

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

Given Function:

$f(x) = \ln(x - 5)$

Given Interval:

[ [a, b] = [6, 14] ]

Step 1: Check Conditions for MVT

• The function $f(x) = \ln(x - 5)$ is defined and continuous for $x > 5$, and differentiable for $x > 5$.
• Since the given interval $[6, 14]$ is completely within this domain, the function satisfies the conditions of the MVT.

Step 2: Compute $f'(x)$

$f'(x) = \frac{1}{x - 5}$

Step 3: Compute the Average Rate of Change

$\frac{f(b) - f(a)}{b - a} = \frac{\ln(14 - 5) - \ln(6 - 5)}{14 - 6}$

$= \frac{\ln 9 - \ln 1}{8}$

Since $\ln 1 = 0$, we get:

$= \frac{\ln 9}{8}$

Step 4: Solve for $c$

Using MVT:

$\frac{1}{c - 5} = \frac{\ln 9}{8}$

Solving for $c$:

$c - 5 = \frac{8}{\ln 9}$

$c = 5 + \frac{8}{\ln 9}$

Final Answer:

$c = 5 + \frac{8}{\ln 9}$

Would you like a numerical approximation or a more detailed breakdown?

Related Questions:

1. How do you prove the Mean Value Theorem?
2. What are the conditions under which MVT does not apply?
3. Can MVT be used on a function with a discontinuity? Why or why not?
4. How does MVT relate to Rolle's Theorem?
5. Can we apply MVT to absolute value functions? Why or why not?

Tip:

The Mean Value Theorem guarantees at least one $c$, but there could be multiple values of $c$ that satisfy the condition. Always check for all possible solutions!