The minimum possible value of $\frac{x^2-6x+10}{3-x}$, for $x < 3$, is

We need to find the minimum value of $\frac{x^2-6x+10}{3-x}$ when $x < 3$.

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Let us rewrite the numerator $x^2 - 6x + 10$ by completing the square.

For any quadratic $x^2 - 6x$, we take half of the coefficient of $x$ and square it: $\left(\frac{6}{2}\right)^2 = 9$

So: $x^2 - 6x + 10 = x^2 - 6x + 9 - 9 + 10 = (x-3)^2 + 1$

Why does this work? Because $(x-3)^2 = x^2 - 6x + 9$, so we add and subtract 9 to complete the square.

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$\frac{x^2-6x+10}{3-x} = \frac{(x-3)^2 + 1}{3-x}$

Now here's a key insight: since $(x-3)^2 = (3-x)^2$ (squaring eliminates the negative sign), we can write:

$\frac{(x-3)^2 + 1}{3-x} = \frac{(3-x)^2 + 1}{3-x}$

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$\frac{(3-x)^2 + 1}{3-x} = \frac{(3-x)^2}{3-x} + \frac{1}{3-x} = (3-x) + \frac{1}{3-x}$

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Since $x < 3$, we have $3-x > 0$. Let us substitute $y = 3-x$ where $y > 0$.

The expression becomes: $y + \frac{1}{y}$

Finding the minimum of $y + \frac{1}{y}$ when $y > 0$:

We'll use the AM-GM (Arithmetic Mean ≥ Geometric Mean) inequality:

$\frac{y + \frac{1}{y}}{2} \geq \sqrt{y \cdot \frac{1}{y}} = \sqrt{1} = 1$

Therefore: $y + \frac{1}{y} \geq 2$

When does equality occur? When $y = \frac{1}{y}$, which means $y^2 = 1$, so $y = 1$ (since $y > 0$).

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When $y = 1$: $3-x = 1$, so $x = 2$

Let us verify: $x = 2 < 3$

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Therefore, the minimum possible value is 2.

Key takeaway: When you see an expression like $\frac{\text{quadratic}}{\text{linear}}$ where the constraint makes the denominator positive, try completing the square in the numerator and look for opportunities to apply AM-GM inequality.