The Mean Value Theorem — Question 2

Question 2 - Solution

We are given: $$f(x) = \ln(x)$$

(a) Check MVT hypotheses:

1. $f(x)$ is continuous on $[1, 4]$: Yes, $\ln(x)$ is continuous for $x > 0$, so continuous on $[1, 4]$

2. $f(x)$ is differentiable on $(1, 4)$: Yes, $\ln(x)$ is differentiable for $x > 0$, so differentiable on $(1, 4)$

✅ The Mean Value Theorem applies.

(b) Find the value(s) of $c$:

Compute average rate of change: $$\frac{f(4) - f(1)}{4 - 1} = \frac{\ln(4) - \ln(1)}{3} = \frac{\ln(4)}{3} = \frac{2\ln(2)}{3}$$

Now compute $f'(x)$: $$f'(x) = \frac{1}{x}$$

Set: $$f'(c) = \frac{1}{c} = \frac{2\ln(2)}{3} \quad \Rightarrow \quad c = \frac{3}{2\ln(2)}$$

Answer: $$\boxed{c = \frac{3}{2\ln(2)} \approx 2.164}$$

Graph of $f(x) = \ln(x)$ on $[1, 4]$: