Verify the truth of Rolle's Theorem for the following functions
`f(x)=sqrt(x-2)" in "[1,2]`

Answer

To verify the truth of Rolle's Theorem for the function \( f(x) = \sqrt{x - 2} \) on the interval \([1, 2]\), we will follow these steps:

### Step 1: Check the continuity of \( f(x) \) on the closed interval \([1, 2]\).

The function \( f(x) = \sqrt{x - 2} \) is defined when \( x - 2 \geq 0 \), which implies \( x \geq 2 \). Therefore, \( f(x) \) is defined only for \( x \) values greater than or equal to 2.

- At \( x = 1 \): \( f(1) = \sqrt{1 - 2} = \sqrt{-1} \) (not defined)
- At \( x = 2 \): \( f(2) = \sqrt{2 - 2} = \sqrt{0} = 0 \)

Since \( f(x) \) is not defined at \( x = 1 \), it is not continuous on the closed interval \([1, 2]\).

### Step 2: Check the differentiability of \( f(x) \) on the open interval \( (1, 2) \).

For \( f(x) \) to be differentiable, it must be defined in the open interval \( (1, 2) \). However, as established in Step 1, \( f(x) \) is not defined for \( x < 2 \). Therefore, \( f(x) \) is not differentiable on the open interval \( (1, 2) \).

### Step 3: Check if \( f(a) = f(b) \).

Here, \( a = 1 \) and \( b = 2 \):
- \( f(1) \) is not defined.
- \( f(2) = 0 \).

Since \( f(1) \) is not defined, we cannot check if \( f(1) = f(2) \).

### Conclusion

Since the function \( f(x) = \sqrt{x - 2} \) is not continuous on the closed interval \([1, 2]\) and not differentiable on the open interval \((1, 2)\), and \( f(1) \) is not defined, we conclude that Rolle's Theorem is not applicable for this function on the given interval.