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.

Explanation

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

### Step 1: Verify the conditions of Rolle's Theorem
Rolle's theorem states that if a function \( f(x) \) 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 \( f'(c) = 0 \).

Given the function \( y = x(x - 4) \), we can rewrite it as:
\[ f(x) = x^2 - 4x \]

We need to check the following:
- **Continuity**: Since \( f(x) \) is a polynomial, it is continuous everywhere, including the interval \([0, 4]\).
- **Differentiability**: Since \( f(x) \) is a polynomial, it is also differentiable everywhere, including the interval \((0, 4)\).
- **Equal values at endpoints**: We need to check if \( f(0) = f(4) \).

Calculating the values:
\[ f(0) = 0(0 - 4) = 0 \]
\[ f(4) = 4(4 - 4) = 0 \]
Since \( f(0) = f(4) = 0 \), the conditions for Rolle's theorem are satisfied.

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

### Step 3: Set the derivative equal to zero
To find the point where the tangent is parallel to the x-axis, we set the derivative equal to zero:
\[ 2x - 4 = 0 \]

### Step 4: Solve for \( x \)
Solving the equation gives:
\[ 2x = 4 \implies x = 2 \]

### Step 5: Find the corresponding \( y \) value
Now we need to find the \( y \) value when \( x = 2 \):
\[ y = f(2) = 2(2 - 4) = 2 \times -2 = -4 \]

### Step 6: Write the coordinates of the point
Thus, the point on the curve where the tangent is parallel to the x-axis is:
\[ (2, -4) \]

### Final Answer
The point on the curve \( y = x(x - 4) \) where the tangent is parallel to the x-axis is \( (2, -4) \).

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