The question in the image asks whether the function $f(x) = \ln(x)$ satisfies the hypotheses of the Mean Value Theorem (MVT) on the interval $[1, 3]$, and if so, to find all numbers $c$ that satisfy the conclusion of the theorem.

Step 1: Verify the Hypotheses of the Mean Value Theorem

The Mean Value Theorem states that if a function $f(x)$:

1. Is continuous on the closed interval $[a, b]$, and
2. Is differentiable on the open interval $(a, b)$,

then there exists at least one number $c \in (a, b)$ such that: $f'(c) = \frac{f(b) - f(a)}{b - a}.$

Verify Continuity and Differentiability:

• The natural logarithmic function $\ln(x)$ is defined, continuous, and differentiable for all $x > 0$.
• On the interval $[1, 3]$, $\ln(x)$ is continuous on $[1, 3]$ and differentiable on $(1, 3)$.

Thus, the hypotheses of the MVT are satisfied.

Step 2: Find $c$ That Satisfies the Conclusion

The conclusion of the MVT states: $f'(c) = \frac{f(3) - f(1)}{3 - 1}.$

Compute the Slope of the Secant Line:

$f(3) = \ln(3), \quad f(1) = \ln(1) = 0.$ $\frac{f(3) - f(1)}{3 - 1} = \frac{\ln(3) - 0}{2} = \frac{\ln(3)}{2}.$

Compute $f'(c)$:

The derivative of $f(x) = \ln(x)$ is: $f'(x) = \frac{1}{x}.$

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

Solve for $c$: $c = \frac{2}{\ln(3)}.$

Final Answer:

The value of $c$ is: $c = \frac{2}{\ln(3)}.$

Let me know if you want additional details or explanations!

Related Questions:

1. What is the geometric interpretation of the Mean Value Theorem?
2. Why is differentiability a necessary condition for the MVT?
3. How do you apply the Mean Value Theorem in practical scenarios?
4. Can $f(x) = |x|$ satisfy the MVT? Why or why not?
5. What happens if the function $f(x)$ is not continuous on the interval?

Tip:

For any interval, always check continuity and differentiability first before applying the MVT!