`f(x) = x(x-1)^(2)` in `[0,1]`

Explanation

To verify Rolle's theorem for the function \( f(x) = x(x-1)^2 \) on the interval \([0, 1]\), we will follow these steps:

### Step 1: Check the continuity of \( f(x) \) on \([0, 1]\)
Since \( f(x) \) is a polynomial function, it is continuous everywhere, including the closed interval \([0, 1]\).

### Step 2: Check the differentiability of \( f(x) \) on \((0, 1)\)
Polynomial functions are also differentiable everywhere, so \( f(x) \) is differentiable on the open interval \((0, 1)\).

### Step 3: Evaluate \( f(0) \) and \( f(1) \)
Now we compute the values of the function at the endpoints of the interval:
- \( f(0) = 0(0-1)^2 = 0 \)
- \( f(1) = 1(1-1)^2 = 0 \)

### Step 4: Check if \( f(0) = f(1) \)
Since \( f(0) = 0 \) and \( f(1) = 0 \), we have \( f(0) = f(1) \).

### Step 5: Apply Rolle's Theorem
Since \( f(x) \) is continuous on \([0, 1]\), differentiable on \((0, 1)\), and \( f(0) = f(1) \), we can apply Rolle's theorem. According to Rolle's theorem, there exists at least one \( c \) in the interval \((0, 1)\) such that \( f'(c) = 0 \).

### Step 6: Find the derivative \( f'(x) \)
Using the product rule, we differentiate \( f(x) \):
\[ f'(x) = \frac{d}{dx}[x] \cdot (x-1)^2 + x \cdot \frac{d}{dx}[(x-1)^2] \]
Calculating the derivatives:
- The derivative of \( x \) is \( 1 \).
- The derivative of \( (x-1)^2 \) is \( 2(x-1) \).

Thus,
\[ f'(x) = (x-1)^2 + x \cdot 2(x-1) = (x-1)^2 + 2x(x-1) \]
Now, simplifying:
\[ f'(x) = (x-1)^2 + 2x^2 - 2x = (x^2 - 2x + 1) + (2x^2 - 2x) = 3x^2 - 4x + 1 \]

### Step 7: Set \( f'(c) = 0 \)
Now we need to solve the equation:
\[ 3c^2 - 4c + 1 = 0 \]

### Step 8: Solve the quadratic equation
Using the quadratic formula \( c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 3 \), \( b = -4 \), and \( c = 1 \).
\[ c = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 3 \cdot 1}}{2 \cdot 3} = \frac{4 \pm \sqrt{16 - 12}}{6} = \frac{4 \pm 2}{6} \]
Calculating the roots:
1. \( c = \frac{6}{6} = 1 \)
2. \( c = \frac{2}{6} = \frac{1}{3} \)

### Step 9: Identify valid \( c \) in \((0, 1)\)
The values of \( c \) are \( \frac{1}{3} \) and \( 1 \). However, \( c \) must be in the open interval \((0, 1)\), so we take \( c = \frac{1}{3} \).

### Conclusion
Thus, we have verified that there exists a \( c \in (0, 1) \) such that \( f'(c) = 0 \), confirming that Rolle's theorem holds for the function \( f(x) = x(x-1)^2 \) on the interval \([0, 1]\).

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