Using Rolle's theroem, find the point on the curve `y = x (x-4), x in [0,4]`, where the tangent is parallel to X-axis.

Answer

To solve the problem using Rolle's theorem, we will follow these steps:

### Step 1: Define the function
We are given the function:
\[ f(x) = x(x - 4) \]
This can also be expressed as:
\[ f(x) = x^2 - 4x \]

### Step 2: 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 point \( c \) in \((a, b)\) such that \( f'(c) = 0 \).

1. **Continuity**: Since \( f(x) \) is a polynomial, it is continuous everywhere, including the interval \([0, 4]\).
2. **Differentiability**: Polynomials are also differentiable everywhere, including the open interval \((0, 4)\).
3. **Equal values at endpoints**:
- Calculate \( f(0) \):
\[ f(0) = 0(0 - 4) = 0 \]
- Calculate \( f(4) \):
\[ f(4) = 4(4 - 4) = 0 \]
- Since \( f(0) = f(4) = 0 \), the conditions of Rolle's theorem are satisfied.

### Step 3: Find the derivative of the function
Now we need to find the derivative \( f'(x) \):
\[ f'(x) = \frac{d}{dx}(x^2 - 4x) = 2x - 4 \]

### Step 4: Set the derivative equal to zero
According to Rolle's theorem, we need to find \( c \) such that:
\[ f'(c) = 0 \]
So we set the derivative equal to zero:
\[ 2c - 4 = 0 \]

### Step 5: Solve for \( c \)
Solving the equation:
\[ 2c - 4 = 0 \]
\[ 2c = 4 \]
\[ c = 2 \]

### Step 6: Verify that \( c \) is in the interval
The value \( c = 2 \) is indeed in the open interval \((0, 4)\).

### Step 7: Find the corresponding \( y \)-coordinate
Now we need to find the \( y \)-coordinate of the point on the curve where the tangent is parallel to the x-axis:
\[ f(2) = 2(2 - 4) = 2(-2) = -4 \]

### Conclusion
The point on the curve where the tangent is parallel to the x-axis is:
\[ (2, -4) \]