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We are given the function \( f(x) \), which is defined piecewise as follows:

\[
f(x) = 
\begin{cases} 
3 - x^2 & \text{if } x \leq 1 \\
\frac{2}{x} & \text{if } x > 1 
\end{cases}
\]
on the interval \( [0, 2] \).

We are tasked with:
1. Verifying that the hypothesis of the Mean Value Theorem (MVT) is satisfied.
2. Finding a suitable choice for \( c \) that satisfies the conclusion of the Mean Value Theorem.

### 1. Verify the hypotheses of the Mean Value Theorem (MVT)

The MVT has two conditions that must be satisfied:
- **Continuity** on the closed interval \([0, 2]\).
- **Differentiability** on the open interval \((0, 2)\).

#### Checking continuity on \([0, 2]\):
- \( f(x) = 3 - x^2 \) is a polynomial and thus continuous for \( x \leq 1 \).
- \( f(x) = \frac{2}{x} \) is continuous for \( x > 1 \).
  
At \( x = 1 \), we must check if the function is continuous at the boundary between the two cases:
\[
\lim_{x \to 1^-} f(x) = 3 - (1)^2 = 2
\]
\[
\lim_{x \to 1^+} f(x) = \frac{2}{1} = 2
\]
Since \( f(1) = 2 \), the function is continuous at \( x = 1 \), and hence continuous on \([0, 2]\).

#### Checking differentiability on \((0, 2)\):
- \( f(x) = 3 - x^2 \) is differentiable on \( (0, 1) \).
- \( f(x) = \frac{2}{x} \) is differentiable on \( (1, 2) \).
  
At \( x = 1 \), we must check if the derivatives from both sides are equal for differentiability:
- The derivative of \( f(x) = 3 - x^2 \) is \( f'(x) = -2x \). At \( x = 1 \), this gives \( f'(1^-) = -2 \).
- The derivative of \( f(x) = \frac{2}{x} \) is \( f'(x) = -\frac{2}{x^2} \). At \( x = 1 \), this gives \( f'(1^+) = -2 \).

Since the left and right derivatives are equal at \( x = 1 \), the function is differentiable on \( (0, 2) \).

Thus, both conditions of the MVT are satisfied.

### 2. Find a suitable choice for \( c \)

The Mean Value Theorem states that there exists some \( c \in (0, 2) \) such that:
\[
f'(c) = \frac{f(2) - f(0)}{2 - 0}
\]

First, we calculate \( f(0) \) and \( f(2) \):
\[
f(0) = 3 - (0)^2 = 3
\]
\[
f(2) = \frac{2}{2} = 1
\]
Thus, the average rate of change is:
\[
\frac{f(2) - f(0)}{2 - 0} = \frac{1 - 3}{2 - 0} = \frac{-2}{2} = -1
\]

Now, we need to find \( c \in (0, 2) \) such that \( f'(c) = -1 \).

- For \( x \in (0, 1) \), \( f'(x) = -2x \). Setting \( -2x = -1 \), we get \( x = \frac{1}{2} \), which lies in the interval \( (0, 1) \).
  
- For \( x \in (1, 2) \), \( f'(x) = -\frac{2}{x^2} \). Setting \( -\frac{2}{x^2} = -1 \), we get \( x^2 = 2 \), so \( x = \sqrt{2} \), which lies in the interval \( (1, 2) \).

Thus, the suitable choices for \( c \) are \( c = \frac{1}{2} \) and \( c = \sqrt{2} \).

---

Would you like further details on any step?

Here are five related questions for exploration:
1. What are other conditions where the Mean Value Theorem is not applicable?
2. How do you interpret the geometric meaning of the Mean Value Theorem?
3. Can the Mean Value Theorem be applied to functions with discontinuities?
4. How can the Mean Value Theorem help find intervals of increasing or decreasing behavior?
5. What are the differences between Rolle’s Theorem and the Mean Value Theorem?

**Tip:** Always check both continuity and differentiability when applying the MVT, as one or both may fail at critical points!

Given the function f(x) = {3 - x^2 for x <= 1, 2/x for x > 1 on [0, 2], verify if the hypothesis of the Mean Value Theorem is satisfied. Find a suitable c that satisfies the conclusion of the theorem.

Verify the hypothesis of the Mean Value Theorem for a piecewise function defined as f(x) = 3 - x^2 for x <= 1 and f(x) = 2/x for x > 1 on [0, 2]. Find a suitable c value that satisfies the conclusion of the MVT. This problem covers the concepts of continuity, differentiability, and the Mean Value Theorem for Grades 10-12 and college students.

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