In this problem, we are working with the function $f(x) = |x - 2|$, and the goal is to analyze whether there exists a point $c$ in the interval $(1,3)$ that satisfies the Mean Value Theorem (MVT) for this function. Let’s go through the steps and explanations provided in your work.

Part (a) Analysis:

1. Determine Continuity and Differentiability:

   • $f(x) = |x - 2|$ is indeed continuous on the interval $[1, 3]$ because the absolute value function is continuous everywhere.
   • However, $f(x) = |x - 2|$ is not differentiable at $x = 2$, as there is a sharp corner (or cusp) at this point. The left-hand derivative and right-hand derivative at $x = 2$ do not match.

2. Calculate $f(3) - f(1)$ and the Slope of the Secant Line:

   • From your work: $f(3) = |3 - 2| = 1$ $f(1) = |1 - 2| = 1$
   • Therefore, the slope of the secant line from $x = 1$ to $x = 3$ is: $\frac{f(3) - f(1)}{3 - 1} = \frac{1 - 1}{3 - 1} = \frac{0}{2} = 0$

3. Mean Value Theorem Requirement:

   • The Mean Value Theorem states that if $f(x)$ is continuous on $[1,3]$ and differentiable on $(1,3)$, then there exists some $c \in (1,3)$ such that: $f'(c) = \frac{f(3) - f(1)}{3 - 1} = 0$
   • However, since $f(x) = |x - 2|$ is not differentiable at $x = 2$, there is no point $c$ in $(1,3)$ where $f'(c) = 0$. The function does not meet the MVT requirements of differentiability over the entire open interval $(1,3)$.

Part (b) Explanation:

The explanation in your work clarifies why this does not contradict the Mean Value Theorem. Specifically, the MVT does not apply here because the function $f(x) = |x - 2|$ is not differentiable at $x = 2$. The theorem requires differentiability on the open interval $(1,3)$, which is not satisfied due to the non-differentiable point at $x = 2$.

Summary:

1. Continuity: $f(x) = |x - 2|$ is continuous on $[1,3]$.
2. Differentiability: $f(x) = |x - 2|$ is not differentiable at $x = 2$.
3. Secant Line Slope: The slope of the secant line is $0$, but there is no $c$ in $(1,3)$ with $f'(c) = 0$ due to the lack of differentiability at $x = 2$.
4. Conclusion: This does not contradict the Mean Value Theorem because the function does not meet the differentiability requirement over $(1,3)$.

Would you like further details on this explanation or help with another question?

Further Questions

1. How would you prove the differentiability of a function at a specific point?
2. Can the Mean Value Theorem apply to functions with discontinuities?
3. How can you determine if a function is differentiable on a given interval?
4. What are the conditions under which the Mean Value Theorem applies?
5. How does the Mean Value Theorem relate to Rolle's Theorem?

Tip:

When applying the Mean Value Theorem, always ensure the function meets both continuity on the closed interval and differentiability on the open interval conditions.