Verify the truth of Rolle's Theorem for the following functions
`f(x)=[x]" in "[-1,1]`

Answer

To verify the truth of Rolle's Theorem for the function \( f(x) = [x] \) (the greatest integer function) on the interval \([-1, 1]\), we need to check the conditions of Rolle's Theorem:

1. The function \( f(x) \) must be continuous on the closed interval \([-1, 1]\).
2. The function \( f(x) \) must be differentiable on the open interval \((-1, 1)\).
3. The function values at the endpoints must be equal, i.e., \( f(-1) = f(1) \).

Let's go through these steps one by one.

### Step 1: Check if \( f(x) \) is continuous on \([-1, 1]\)

The greatest integer function \( f(x) = [x] \) is defined as the largest integer less than or equal to \( x \).

- For \( x \) in the interval \([-1, 0)\), \( f(x) = -1 \).
- For \( x \) in the interval \([0, 1)\), \( f(x) = 0 \).
- At \( x = -1 \), \( f(-1) = -1 \).
- At \( x = 1 \), \( f(1) = 1 \).

The function has discontinuities at integer points. Specifically, at \( x = 0 \) and \( x = 1 \), the function jumps from -1 to 0 and from 0 to 1, respectively. Therefore, \( f(x) \) is not continuous at \( x = 1 \).

### Step 2: Check if \( f(x) \) is differentiable on \((-1, 1)\)

Since the function is not continuous at \( x = 1 \), it cannot be differentiable at that point. Furthermore, the greatest integer function is not differentiable at any integer point. Thus, \( f(x) \) is not differentiable on \((-1, 1)\).

### Step 3: Check if \( f(-1) = f(1) \)

We have:
- \( f(-1) = -1 \)
- \( f(1) = 1 \)

Clearly, \( f(-1) \neq f(1) \).

### Conclusion

Since the function \( f(x) = [x] \) is not continuous on \([-1, 1]\) and is not differentiable on \((-1, 1)\), and the function values at the endpoints are not equal, we conclude that the conditions of Rolle's Theorem are not satisfied.

Thus, we can say that Rolle's Theorem is not applicable for the function \( f(x) = [x] \) on the interval \([-1, 1]\).

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