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 for the given function \( f(x) \), we need to follow these steps:

### Step 1: Define the function
The function \( f(x) \) 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 continuity
To apply Rolle's theorem, the function must be continuous on the closed interval \([0, 2]\). We need to check the continuity at the point where the definition of the function changes, which is at \( x = 1 \).

1. Calculate \( f(1) \):
\[ f(1) = 1^2 + 1 = 2 \]

2. Calculate the left-hand limit as \( x \) approaches 1:
\[ \lim_{x \to 1^-} f(x) = f(1) = 2 \]

3. Calculate the right-hand limit as \( x \) approaches 1:
\[ \lim_{x \to 1^+} f(x) = 3 - 1 = 2 \]

Since both the left-hand limit and right-hand limit at \( x = 1 \) are equal to \( f(1) \), we conclude that \( f(x) \) is continuous on \([0, 2]\).

### Step 3: Check differentiability
Next, we need to check if \( f(x) \) is differentiable on the open interval \( (0, 2) \). We specifically need to check the differentiability at \( x = 1 \).

1. Calculate the left-hand derivative at \( x = 1 \):
\[ f'(1^-) = \lim_{h \to 0} \frac{f(1) - f(1-h)}{h} \]
Here, \( f(1-h) = (1-h)^2 + 1 = 1 - 2h + h^2 + 1 = 2 - 2h + h^2 \).
\[ f'(1^-) = \lim_{h \to 0} \frac{2 - (2 - 2h + h^2)}{h} = \lim_{h \to 0} \frac{2h - h^2}{h} = \lim_{h \to 0} (2 - h) = 2 \]

2. Calculate the right-hand derivative at \( x = 1 \):
\[ f'(1^+) = \lim_{h \to 0} \frac{f(1+h) - f(1)}{h} \]
Here, \( f(1+h) = 3 - (1+h) = 2 - h \).
\[ f'(1^+) = \lim_{h \to 0} \frac{(2 - h) - 2}{h} = \lim_{h \to 0} \frac{-h}{h} = -1 \]

Since the left-hand derivative \( f'(1^-) = 2 \) and the right-hand derivative \( f'(1^+) = -1 \) are not equal, \( f(x) \) is not differentiable at \( x = 1 \).

### Step 4: Conclusion
Since \( f(x) \) is continuous on \([0, 2]\) but not differentiable at \( x = 1\), we conclude that Rolle's theorem is not applicable for the function \( f(x) \) on the interval \([0, 2]\).

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