Verify Rolle's Theorem for the function :
`f(x)={{:(-4x+5", "0lexle1),(2x-3" , "1ltxle2):}`

Answer

To verify Rolle's Theorem for the given piecewise function \( f(x) \), we will follow these steps:

### Step 1: Define the function and intervals
The function is defined as:
\[ f(x) = \begin{cases} -4x + 5 & \text{for } 0 \leq x \leq 1 \\ 2x - 3 & \text{for } 1 < x \leq 2 \end{cases} \]
We need to check the intervals \( [0, 1] \) and \( [1, 2] \).

### Step 2: Check continuity on the closed intervals
1. **For the interval \( [0, 1] \)**:
- The function \( f(x) = -4x + 5 \) is a polynomial, which is continuous everywhere. Thus, \( f(x) \) is continuous on \( [0, 1] \).

2. **For the interval \( [1, 2] \)**:
- The function \( f(x) = 2x - 3 \) is also a polynomial, hence continuous everywhere. Thus, \( f(x) \) is continuous on \( [1, 2] \).

### Step 3: Check differentiability on the open intervals
1. **For the interval \( (0, 1) \)**:
- The function \( f(x) = -4x + 5 \) is differentiable everywhere, so it is differentiable on \( (0, 1) \).

2. **For the interval \( (1, 2) \)**:
- The function \( f(x) = 2x - 3 \) is also differentiable everywhere, so it is differentiable on \( (1, 2) \).

### Step 4: Check the values at the endpoints
1. **For the interval \( [0, 1] \)**:
- Calculate \( f(0) \):
\[ f(0) = -4(0) + 5 = 5 \]
- Calculate \( f(1) \):
\[ f(1) = -4(1) + 5 = 1 \]
- Since \( f(0) \neq f(1) \) (i.e., \( 5 \neq 1 \)), the condition \( f(a) = f(b) \) is not satisfied.

2. **For the interval \( [1, 2] \)**:
- Calculate \( f(1) \):
\[ f(1) = 2(1) - 3 = -1 \]
- Calculate \( f(2) \):
\[ f(2) = 2(2) - 3 = 1 \]
- Since \( f(1) \neq f(2) \) (i.e., \( -1 \neq 1 \)), the condition \( f(a) = f(b) \) is not satisfied.

### Conclusion
Since the conditions of Rolle's Theorem are not satisfied in both intervals \( [0, 1] \) and \( [1, 2] \), we conclude that:
\[ \text{Rolle's Theorem is not applicable for the function } f(x). \]