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Algebra Difficulty 3.8 AMC 10/12 Find the answer China

The minimum value of f(x)=5−4x+x22−xf(x) = \frac{5-4x+x^2}{2-x} for x∈(−∞,2)x \in (-\infty, 2) is ( ).

This was a multiple-choice question, but the options didn't survive into the source we have. The answer given is C, and the solution below works it through.

Solution

Let x<2⇒2−x>0x < 2 \Rightarrow 2-x > 0. Then
f(x)=1+(4−4x+x2)2−x=12−x+(2−x)≥2×12−x×(2−x)=2. \begin{aligned} f(x) &= \frac{1 + (4 - 4x + x^2)}{2-x} = \frac{1}{2-x} + (2-x) \\ &\geq 2 \times \sqrt{\frac{1}{2-x} \times (2-x)} = 2. \end{aligned}
The equality holds if and only if 12−x=2−x\frac{1}{2-x} = 2-x, and it is so when x=1∈(−∞,2)x = 1 \in (-\infty, 2). This means that ff reaches the minimum value 2 at x=1x = 1.

Answer: C

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