Discuss the applicability of Rolle's theorem on the function given by
`f(x) = {{:(x^(2)+1,if0lexle1),(3-x,if1lexle2):}`

Explanation

To discuss the applicability of Rolle's theorem on the given function \( f(x) \), we need to follow these steps:

### Step 1: Define the function
The function is defined piecewise as follows:
\[ f(x) = \begin{cases} x^2 + 1 & \text{if } 0 \leq x \leq 1 \\ 3 - x & \text{if } 1 < x \leq 2 \end{cases} \]

### Step 2: Check the interval
Rolle's theorem applies to a function that is continuous on a closed interval \([a, b]\) and differentiable on the open interval \((a, b)\). Here, the interval is \([0, 2]\).

### Step 3: Check continuity at \( x = 1 \)
To check continuity at \( x = 1 \), we need to ensure that:
\[ f(1^-) = f(1^+) \]
Calculating \( f(1^-) \):
\[ f(1^-) = 1^2 + 1 = 2 \]
Calculating \( f(1^+) \):
\[ f(1^+) = 3 - 1 = 2 \]
Since \( f(1^-) = f(1^+) = 2 \), the function is continuous at \( x = 1 \).

### Step 4: Check differentiability at \( x = 1 \)
Next, we need to check the differentiability at \( x = 1 \) by finding the left-hand derivative and the right-hand derivative.

**Left-hand derivative at \( x = 1 \)**:
\[ f'(x) = 2x \quad \text{for } 0 \leq x < 1 \]
Thus,
\[ f'(1^-) = 2 \cdot 1 = 2 \]

**Right-hand derivative at \( x = 1 \)**:
\[ f'(x) = -1 \quad \text{for } 1 < x \leq 2 \]
Thus,
\[ f'(1^+) = -1 \]

### Step 5: Compare the derivatives
Now we compare the left-hand and right-hand derivatives:
\[ f'(1^-) = 2 \quad \text{and} \quad f'(1^+) = -1 \]
Since \( f'(1^-) \neq f'(1^+) \), the function is not differentiable at \( x = 1 \).

### Conclusion
Since the function is continuous on the interval \([0, 2]\) but not differentiable at \( x = 1\), we conclude that Rolle's theorem does not apply to the function \( f(x) \).

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