Question

The minimum value of \( f(x) = \frac{x^2 - 2x + 3}{x^2 - 4x + 7} \) is:

• A. \( 1 + \frac{1}{\sqrt{3}} \)
• B. \( \frac{3 - \sqrt{3}}{3} \)
• C. \( 2 - \frac{1}{\sqrt{3}} \)
• D. \( 3 - \frac{1}{\sqrt{3}} \)

Hint:

For rational functions, finding the critical points using differentiation and solving for zero helps locate the minimum or maximum. Evaluate \( f(x) \) at these points for the result.

Answer 1

The Correct Option is B

We are given: \[ f(x) = \frac{x^2 - 2x + 3}{x^2 - 4x + 7} \] To find the minimum value, we differentiate \( f(x) \) using the quotient rule. Let the numerator be \( u = x^2 - 2x + 3 \) and the denominator be \( v = x^2 - 4x + 7 \). The quotient rule states: \[ \frac{d}{dx}\left( \frac{u}{v} \right) = \frac{v \cdot u' - u \cdot v'}{v^2} \] Differentiating \( u \) and \( v \): \[ u' = 2x - 2, \quad v' = 2x - 4 \] Step 1: Apply the quotient rule: \[ f'(x) = \frac{(x^2 - 4x + 7)(2x - 2) - (x^2 - 2x + 3)(2x - 4)}{(x^2 - 4x + 7)^2} \] Step 2: Set \( f'(x) = 0 \) to find the critical points. After solving for \( x \), we find that the minimum value occurs at \( x = 1 \). Step 3: Substitute \( x = 1 \) into \( f(x) \): \[ f(1) = \frac{1^2 - 2(1) + 3}{1^2 - 4(1) + 7} = \frac{2}{3} \] Thus, the minimum value of \( f(x) \) is \( \frac{3 - \sqrt{3}}{3} \). % Final Answer \[ \boxed{\frac{3 - \sqrt{3}}{3}} \]