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 \[ \boxed{\frac{3 - \sqrt{3}}{3}} \]

Answer 2

Given:
We are asked to find the minimum value of the function:
\[ f(x) = \frac{x^2 - 2x + 3}{x^2 - 4x + 7} \]

Step 1: Simplify using substitution
To simplify, complete the square in both numerator and denominator.
Numerator: \( x^2 - 2x + 3 = (x - 1)^2 + 2 \)
Denominator: \( x^2 - 4x + 7 = (x - 2)^2 + 3 \)

So the function becomes:
\[ f(x) = \frac{(x - 1)^2 + 2}{(x - 2)^2 + 3} \]
Let \( t = x - 2 \Rightarrow x = t + 2 \)
Then \( x - 1 = t + 1 \), and the function becomes:
\[ f(t) = \frac{(t + 1)^2 + 2}{t^2 + 3} = \frac{t^2 + 2t + 1 + 2}{t^2 + 3} = \frac{t^2 + 2t + 3}{t^2 + 3} \]

Step 2: Let’s simplify further
\[ f(t) = \frac{t^2 + 3 + 2t}{t^2 + 3} = 1 + \frac{2t}{t^2 + 3} \]
Now, define: \( f(t) = 1 + \frac{2t}{t^2 + 3} \)

Step 3: Minimize the function
Let \( y = 1 + \frac{2t}{t^2 + 3} \). We now minimize this expression.

Let’s define \( g(t) = \frac{2t}{t^2 + 3} \)
Differentiate \( g(t) \):
\[ g'(t) = \frac{2(t^2 + 3) - 2t(2t)}{(t^2 + 3)^2} = \frac{2(t^2 + 3 - 2t^2)}{(t^2 + 3)^2} = \frac{2(3 - t^2)}{(t^2 + 3)^2} \]
Set \( g'(t) = 0 \Rightarrow 3 - t^2 = 0 \Rightarrow t^2 = 3 \Rightarrow t = \pm\sqrt{3} \)

Step 4: Evaluate minimum value
Substitute \( t = -\sqrt{3} \) into \( f(t) = 1 + \frac{2t}{t^2 + 3} \):
\[ f(-\sqrt{3}) = 1 + \frac{2(-\sqrt{3})}{3 + 3} = 1 - \frac{\sqrt{3}}{3} = \frac{3 - \sqrt{3}}{3} \]

Final Answer:
The minimum value of \( f(x) \) is \( \boxed{\frac{3 - \sqrt{3}}{3}} \).