Verify Rolle's Theorem in the interval [a, b] for the function :
`f(x)=(x-a)^(2)(x-b)^(2)`. Find the value of 'c'.

Answer

To verify Rolle's Theorem for the function \( f(x) = (x-a)^2 (x-b)^2 \) in the interval \([a, b]\), we need to follow these steps:

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

1. **Continuity**: The function \( f(x) \) is a polynomial (specifically a product of two quadratic functions). Since polynomials are continuous everywhere, \( f(x) \) is continuous on \([a, b]\).

2. **Differentiability**: Similarly, since \( f(x) \) is a polynomial, it is differentiable everywhere, including on the open interval \((a, b)\).

### Step 2: Check the Values at the Endpoints
Next, we need to check if \( f(a) = f(b) \).

- Calculate \( f(a) \):
\[ f(a) = (a-a)^2 (a-b)^2 = 0 \cdot (a-b)^2 = 0 \]

- Calculate \( f(b) \):
\[ f(b) = (b-a)^2 (b-b)^2 = (b-a)^2 \cdot 0 = 0 \]

Since \( f(a) = 0 \) and \( f(b) = 0 \), we have \( f(a) = f(b) \).

### Step 3: Find \( c \) such that \( f'(c) = 0 \)
Now we need to find the derivative \( f'(x) \) and set it to zero.

Using the product rule:
\[ f'(x) = \frac{d}{dx}[(x-a)^2] \cdot (x-b)^2 + (x-a)^2 \cdot \frac{d}{dx}[(x-b)^2] \]
Calculating the derivatives:
\[ \frac{d}{dx}[(x-a)^2] = 2(x-a) \]
\[ \frac{d}{dx}[(x-b)^2] = 2(x-b) \]
Thus,
\[ f'(x) = 2(x-a)(x-b)^2 + (x-a)^2 \cdot 2(x-b) \]
Factoring out common terms:
\[ f'(x) = 2(x-a)(x-b) \left[(x-b) + (x-a)\right] = 2(x-a)(x-b)(2x - (a+b)) \]

### Step 4: Set \( f'(c) = 0 \)
Setting the derivative equal to zero:
\[ 2(x-a)(x-b)(2x - (a+b)) = 0 \]
This gives us three cases:
1. \( x - a = 0 \) → \( x = a \)
2. \( x - b = 0 \) → \( x = b \)
3. \( 2x - (a+b) = 0 \) → \( x = \frac{a+b}{2} \)

### Step 5: Determine Valid \( c \)
The values \( x = a \) and \( x = b \) are not in the open interval \((a, b)\). However, \( x = \frac{a+b}{2} \) is in the open interval \((a, b)\).

### Conclusion
Thus, we have verified that all conditions of Rolle's Theorem are satisfied, and we found that:
\[ c = \frac{a+b}{2} \]