Verify Rolle's theorem for `f(x)=x,x in[1,2]`.

Answer

To verify Rolle's theorem for the function \( f(x) = x \) on the interval \([1, 2]\), we need to follow these steps:

### Step 1: Check the conditions of Rolle's Theorem
Rolle's theorem states that if a function \( f \) is continuous on the closed interval \([a, b]\), differentiable on the open interval \((a, b)\), and \( f(a) = f(b) \), then there exists at least one \( c \) in \((a, b)\) such that \( f'(c) = 0 \).

### Step 2: Verify continuity
The function \( f(x) = x \) is a polynomial function, which is continuous everywhere. Therefore, it is continuous on the interval \([1, 2]\).

**Hint:** A polynomial function is continuous on its entire domain.

### Step 3: Verify differentiability
The function \( f(x) = x \) is also differentiable everywhere since it is a polynomial function. Thus, it is differentiable on the interval \((1, 2)\).

**Hint:** A polynomial function is differentiable on its entire domain.

### Step 4: Check the values at the endpoints
Now we need to check the values of the function at the endpoints of the interval:
- \( f(1) = 1 \)
- \( f(2) = 2 \)

We see that \( f(1) \neq f(2) \) (specifically, \( f(1) = 1 \) and \( f(2) = 2 \)).

**Hint:** For Rolle's theorem to apply, the function values at the endpoints must be equal.

### Step 5: Conclusion
Since \( f(1) \neq f(2) \), the condition \( f(a) = f(b) \) is not satisfied. Therefore, we cannot apply Rolle's theorem to the function \( f(x) = x \) on the interval \([1, 2]\).

**Final Answer:** Rolle's theorem is not verified for the function \( f(x) = x \) on the interval \([1, 2]\) because \( f(1) \neq f(2) \).