Let: \[ f(x) = \frac{u}{v}, \quad \text{where } u = x^2 - 6x + 10, \quad v = 3 - x \] Using the quotient rule: \[ f'(x) = \frac{v \cdot u' - u \cdot v'}{v^2} \] Compute derivatives: \[ u' = 2x - 6, \quad v' = -1 \] Substituting these into the quotient rule: \[ f'(x) = \frac{(3 - x)(2x - 6) - (x^2 - 6x + 10)(-1)}{(3 - x)^2} \] Simplify the numerator: \[ (3 - x)(2x - 6) + (x^2 - 6x + 10) = (6x - 18 - 2x^2 + 6x) + x^2 - 6x + 10 \] Combining like terms: \[ -2x^2 + 12x - 18 + x^2 - 6x + 10 = -x^2 + 6x - 8 \] Thus, the derivative is: \[ f'(x) = \frac{-x^2 + 6x - 8}{(3 - x)^2} \] To find critical points, set \( f'(x) = 0 \): \[ -x^2 + 6x - 8 = 0 \]
Solving the equation \( -x^2 + 6x - 8 = 0 \) is equivalent to solving \( x^2 - 6x + 8 = 0 \). Using the quadratic formula: \[ x = \frac{6 \pm \sqrt{(-6)^2 - 4(1)(8)}}{2(1)} = \frac{6 \pm \sqrt{36 - 32}}{2} = \frac{6 \pm \sqrt{4}}{2} = \frac{6 \pm 2}{2} \] This gives two potential critical points: \( x = \frac{6 + 2}{2} = 4 \) and \( x = \frac{6 - 2}{2} = 2 \). Given the domain constraint \( x<3 \), only \( x = 2 \) is a valid critical point.
Evaluate the function \( f(x) \) at the valid critical point \( x = 2 \): \[ f(2) = \frac{2^2 - 6 \cdot 2 + 10}{3 - 2} = \frac{4 - 12 + 10}{1} = \frac{2}{1} = 2 \] The minimum value is 2.
The minimum value of the function is 2. This corresponds to Option C.