Discuss the applicability of Rolle's theorem for the function
`f(x)={{:(x^(2)-4",",x le1),(5x-8"," , x gt1):}`

Answer

To discuss the applicability of Rolle's theorem for the given piecewise function

\[ f(x) = \begin{cases} x^2 - 4 & \text{if } x \leq 1 \\ 5x - 8 & \text{if } x > 1 \end{cases} \]

we need to check two main conditions: continuity and differentiability of the function on the interval we are considering.

### Step 1: Check Continuity

To check the continuity of \( f(x) \) at \( x = 1 \), we need to find the left-hand limit (LHL) and right-hand limit (RHL) at that point, and ensure they are equal to the function value at that point.

1. **Left-Hand Limit (LHL) as \( x \to 1^- \)**:
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1} (x^2 - 4) = 1^2 - 4 = 1 - 4 = -3 \]

2. **Right-Hand Limit (RHL) as \( x \to 1^+ \)**:
\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1} (5x - 8) = 5(1) - 8 = 5 - 8 = -3 \]

3. **Function Value at \( x = 1 \)**:
\[ f(1) = 1^2 - 4 = -3 \]

Since LHL = RHL = \( f(1) = -3 \), the function is continuous at \( x = 1 \).

### Step 2: Check Differentiability

Next, we need to check if \( f(x) \) is differentiable at \( x = 1 \).

1. **Derivative for \( x < 1 \)**:
\[ f'(x) = \frac{d}{dx}(x^2 - 4) = 2x \]
Evaluating at \( x = 1 \):
\[ f'(1^-) = 2(1) = 2 \]

2. **Derivative for \( x > 1 \)**:
\[ f'(x) = \frac{d}{dx}(5x - 8) = 5 \]
Evaluating at \( x = 1 \):
\[ f'(1^+) = 5 \]

Since \( f'(1^-) = 2 \) and \( f'(1^+) = 5 \), the left-hand derivative does not equal the right-hand derivative. Therefore, \( f(x) \) is not differentiable at \( x = 1 \).

### Conclusion

Rolle's theorem states that if a function is continuous on a closed interval and differentiable on the open interval, then there exists at least one \( c \) in the open interval such that \( f'(c) = 0 \).

In this case, while the function is continuous on the interval, it is not differentiable at \( x = 1 \). Therefore, **Rolle's theorem does not apply** to this function.