https://file.aiquickdraw.com/mathbot/file/1729231080168ymw4lb94.jpg

Let's analyze the problem step by step.

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

### 1. Verifying the hypothesis of the Mean Value Theorem (MVT):

The Mean Value Theorem requires two conditions:
- **Continuity** on the closed interval \([a, b]\),
- **Differentiability** on the open interval \((a, b)\).

**Continuity:**

- \(f(x) = 3 - x^2\) is a polynomial, and polynomials are continuous. Therefore, \(f(x)\) is continuous on \([0, 1]\).
- \(f(x) = \frac{2}{x}\) is continuous for \(x > 1\), specifically on the interval \((1, 2]\).
- We need to check if the function is continuous at \(x = 1\), where the two pieces of the function meet. We do this by checking the left-hand and right-hand limits at \(x = 1\).

\[
\lim_{x \to 1^-} f(x) = 3 - 1^2 = 2, \quad \lim_{x \to 1^+} f(x) = \frac{2}{1} = 2.
\]
Since both limits are equal and match the function's value at \(x = 1\), the function is continuous at \(x = 1\).

Therefore, \(f(x)\) is continuous on \([0, 2]\).

**Differentiability:**

- \(f(x) = 3 - x^2\) is differentiable for \(x \in (0, 1)\).
- \(f(x) = \frac{2}{x}\) is differentiable for \(x \in (1, 2)\).
- We need to check if \(f(x)\) is differentiable at \(x = 1\). To do this, we check if the left-hand and right-hand derivatives at \(x = 1\) are equal.

For \(x \leq 1\):
\[
f'(x) = \frac{d}{dx}(3 - x^2) = -2x, \quad f'(1) = -2(1) = -2.
\]

For \(x > 1\):
\[
f'(x) = \frac{d}{dx}\left(\frac{2}{x}\right) = -\frac{2}{x^2}, \quad f'(1) = -\frac{2}{1^2} = -2.
\]

Since the left-hand and right-hand derivatives are equal at \(x = 1\), \(f(x)\) is differentiable at \(x = 1\).

Thus, \(f(x)\) is differentiable on \((0, 2)\).

**Conclusion:** Both conditions of the Mean Value Theorem are satisfied.

---

### 2. Finding a suitable \(c\) that satisfies the conclusion of the Mean Value Theorem:

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

Let's calculate the right-hand side:

- \(f(0) = 3 - 0^2 = 3\),
- \(f(2) = \frac{2}{2} = 1\).

Thus, 
\[
\frac{f(2) - f(0)}{2 - 0} = \frac{1 - 3}{2} = \frac{-2}{2} = -1.
\]

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

**For \(x \leq 1\):**
\[
f'(x) = -2x.
\]
We set \(-2x = -1\), which gives \(x = \frac{1}{2}\).

Thus, \(c = \frac{1}{2}\) is a solution in \((0, 1)\).

**For \(x > 1\):**
\[
f'(x) = -\frac{2}{x^2}.
\]
We set \(-\frac{2}{x^2} = -1\), which gives \(x^2 = 2\), or \(x = \sqrt{2} \approx 1.414\).

Thus, \(c = \sqrt{2}\) is a solution in \((1, 2)\).

### Conclusion:
The values \(c = \frac{1}{2}\) and \(c = \sqrt{2}\) both satisfy the conclusion of the Mean Value Theorem.

---

Would you like more details on any of these steps? 

Here are some related questions to expand your understanding:
1. How can we visually interpret the Mean Value Theorem?
2. What happens if a function is not continuous or differentiable at a point on the interval?
3. How does the Mean Value Theorem relate to Rolle's Theorem?
4. Can there be more than one value of \(c\) that satisfies the MVT for a function?
5. What are some applications of the Mean Value Theorem in physics or other fields?

**Tip:** Always check both continuity and differentiability carefully, especially when dealing with piecewise functions, as these are common points where these properties may fail.

Given the function f(x) = {3 - x^2 if x ≤ 1, 2/x if x > 1} on [0,2], verify that the hypothesis of the mean-value theorem is satisfied and find a suitable choice for c that satisfies the conclusion of the mean-value theorem.

Verify the conditions of the Mean Value Theorem and find c for the piecewise function f(x) = {3 - x^2 if x ≤ 1, 2/x if x > 1} on the interval [0, 2]. This detailed explanation covers continuity, differentiability, and applying the MVT to a piecewise function.

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