Discuss the applicability of Rolles theorem on the function `f(x)={x^2+1,\ \ \ \ w h e n\ 0lt=xlt=1 3-x ,\ \ \ \ w h e n\ 1

Explanation

To discuss the applicability of Rolle's theorem on the given function \( f(x) \), we need to check two main conditions:

1. The function must be continuous on the closed interval \([0, 2]\).
2. The function must be differentiable on the open interval \((0, 2)\).

The function is defined as follows:
\[ f(x) = \begin{cases} x^2 + 1 & \text{for } 0 < x < 1 \\ 3 - x & \text{for } 1 < x < 2 \end{cases} \]

### Step 1: Check Continuity

To check the continuity of \( f(x) \) at the point where the definition changes (i.e., at \( x = 1 \)), we need to find the left-hand limit (LHL) and right-hand limit (RHL) at \( x = 1 \).

- **Left-hand limit (LHL) at \( x = 1 \)**:
\[ \text{LHL} = \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^2 + 1) = 1^2 + 1 = 2 \]

- **Right-hand limit (RHL) at \( x = 1 \)**:
\[ \text{RHL} = \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (3 - x) = 3 - 1 = 2 \]

Since LHL = RHL = 2, we conclude that:
\[ f(1) = 2 \]
Thus, \( f(x) \) is continuous at \( x = 1 \).

### Step 2: Check Differentiability

Next, we need to check the differentiability of \( f(x) \) at \( x = 1 \) by finding the left-hand derivative (LHD) and right-hand derivative (RHD).

- **Left-hand derivative (LHD) at \( x = 1 \)**:
\[ \text{LHD} = \lim_{h \to 0^-} \frac{f(1 + h) - f(1)}{h} = \lim_{h \to 0^-} \frac{(1 + h)^2 + 1 - 2}{h} = \lim_{h \to 0^-} \frac{(1 + 2h + h^2) + 1 - 2}{h} = \lim_{h \to 0^-} \frac{2h + h^2}{h} = \lim_{h \to 0^-} (2 + h) = 2 \]

- **Right-hand derivative (RHD) at \( x = 1 \)**:
\[ \text{RHD} = \lim_{h \to 0^+} \frac{f(1 + h) - f(1)}{h} = \lim_{h \to 0^+} \frac{(3 - (1 + h)) - 2}{h} = \lim_{h \to 0^+} \frac{(2 - h) - 2}{h} = \lim_{h \to 0^+} \frac{-h}{h} = -1 \]

Since LHD = 2 and RHD = -1, we find that:
\[ \text{LHD} \neq \text{RHD} \]
Thus, \( f(x) \) is not differentiable at \( x = 1 \).

### Conclusion

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

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