Verify the conditions of Rolle's Theorem in the following problems. In each case, find a point in the interval, where the derivative vanishes :
`x^(2)-1 " on "[-1,1]`

Answer

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

### Step 1: Check Continuity
First, we need to check if the function \( f(x) \) is continuous on the closed interval \([-1, 1]\).

**Solution:**
The function \( f(x) = x^2 - 1 \) is a polynomial function. Polynomial functions are continuous everywhere, including the interval \([-1, 1]\).

### Step 2: Check Differentiability
Next, we need to check if the function \( f(x) \) is differentiable on the open interval \((-1, 1)\).

**Solution:**
Since \( f(x) \) is a polynomial, it is differentiable everywhere, including the open interval \((-1, 1)\).

### Step 3: Check the Values at the Endpoints
Now, we need to check if \( f(-1) = f(1) \).

**Solution:**
Calculate \( f(-1) \) and \( f(1) \):
\[ f(-1) = (-1)^2 - 1 = 1 - 1 = 0 \]
\[ f(1) = (1)^2 - 1 = 1 - 1 = 0 \]
Since \( f(-1) = f(1) = 0 \), the third condition is satisfied.

### Step 4: Find the Derivative
Now we will find the derivative of \( f(x) \).

**Solution:**
The derivative of \( f(x) \) is:
\[ f'(x) = \frac{d}{dx}(x^2 - 1) = 2x \]

### Step 5: Find the Point Where the Derivative Vanishes
We need to find a point \( c \) in the interval \((-1, 1)\) such that \( f'(c) = 0 \).

**Solution:**
Set the derivative equal to zero:
\[ 2c = 0 \]
Solving for \( c \):
\[ c = 0 \]
Since \( c = 0 \) lies within the interval \((-1, 1)\), we have found the point where the derivative vanishes.

### Conclusion
All conditions of Rolle's Theorem are satisfied, and the point where the derivative vanishes is \( c = 0 \).

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