Verify L.M.V. theorem for `f(x)=x(x-2)" on "[1,3]`

Answer

To verify the Lagrange Mean Value Theorem (LMVT) for the function \( f(x) = x(x - 2) \) on the interval \([1, 3]\), we will follow these steps:

### Step 1: Check Continuity and Differentiability

1. **Function Definition**: The function is given as \( f(x) = x(x - 2) = x^2 - 2x \).
2. **Continuity**: Since \( f(x) \) is a polynomial, it is continuous everywhere, including on the interval \([1, 3]\).
3. **Differentiability**: Polynomials are also differentiable everywhere. Thus, \( f(x) \) is differentiable on \([1, 3]\).

### Step 2: Calculate \( f(a) \) and \( f(b) \)

1. **Endpoints**: Let \( a = 1 \) and \( b = 3 \).
2. **Calculate \( f(1) \)**:
\[ f(1) = 1(1 - 2) = 1 \cdot (-1) = -1 \]
3. **Calculate \( f(3) \)**:
\[ f(3) = 3(3 - 2) = 3 \cdot 1 = 3 \]

### Step 3: Apply the Mean Value Theorem Formula

1. **Mean Value Theorem Statement**: According to LMVT, there exists at least one \( c \) in \((a, b)\) such that:
\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]
2. **Calculate \( f(b) - f(a) \)**:
\[ f(b) - f(a) = f(3) - f(1) = 3 - (-1) = 3 + 1 = 4 \]
3. **Calculate \( b - a \)**:
\[ b - a = 3 - 1 = 2 \]
4. **Calculate the right-hand side**:
\[ \frac{f(b) - f(a)}{b - a} = \frac{4}{2} = 2 \]

### Step 4: Find \( f'(x) \)

1. **Differentiate \( f(x) \)**:
\[ f'(x) = \frac{d}{dx}(x^2 - 2x) = 2x - 2 \]

### Step 5: Solve for \( c \)

1. **Set \( f'(c) = 2 \)**:
\[ 2c - 2 = 2 \]
2. **Solve for \( c \)**:
\[ 2c = 4 \implies c = 2 \]

### Step 6: Verify \( c \) is in the interval \((1, 3)\)

1. **Check the interval**: Since \( c = 2 \) is within \((1, 3)\), the conditions of the LMVT are satisfied.

### Conclusion

Since all conditions of the Lagrange Mean Value Theorem are satisfied, we conclude that the theorem holds for the function \( f(x) = x(x - 2) \) on the interval \([1, 3]\).

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