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 will follow these steps:

### Step 1: Define the function
We are given the curve \( y = x(x - 4) \). We can rewrite this as:
\[ y = x^2 - 4x \]
Let \( f(x) = x^2 - 4x \).

### Step 2: Check the conditions of Rolle's theorem
Rolle's theorem states that if a function 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 \( f'(c) = 0 \).

- **Continuity**: The function \( f(x) = x^2 - 4x \) is a polynomial, which is continuous everywhere, including the interval \([0, 4]\).
- **Differentiability**: The function is also differentiable everywhere, including the interval \((0, 4)\).
- **Check endpoints**: We need to check \( f(0) \) and \( f(4) \):
\[ f(0) = 0^2 - 4 \cdot 0 = 0 \]
\[ f(4) = 4^2 - 4 \cdot 4 = 16 - 16 = 0 \]
Since \( f(0) = f(4) = 0 \), the conditions of Rolle's theorem are satisfied.

### Step 3: Find the derivative
Next, we differentiate the function \( f(x) \):
\[ f'(x) = \frac{d}{dx}(x^2 - 4x) = 2x - 4 \]

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

### Step 5: Find the corresponding y-coordinate
Now, we substitute \( x = 2 \) back into the original function to find the corresponding \( y \)-coordinate:
\[ f(2) = 2^2 - 4 \cdot 2 = 4 - 8 = -4 \]

### Step 6: State 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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