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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.

1) Let f(x) = |x - 2|. Show that there is no c in (1, 3) such that (f(3) - f(1)) / (3 - 1) = f'(c), and explain why this does not contradict the Mean Value Theorem.

Analyze the function f(x) = |x - 2| to determine why there is no point c in the interval (1, 3) satisfying the Mean Value Theorem. This problem covers concepts of continuity, differentiability, and the Mean Value Theorem in calculus.

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