# How do you solve ((x-1)(3-x))/(x-2)^2>=0?

Nov 30, 2016

 f(x)≥0 for 1≤x≤3 with the exclusion of $x = 2$ where it is undefined.

#### Explanation:

The denominator is always positive, so the sign of $f \left(x\right)$ is determined by the numerator.

The numerator is positive when both factors are positive or both are negative.

$\left(x - 1\right) \ge 0$ for $x \ge 1$
$\left(3 - x\right) \ge 0$ for $x \le 3$

So $f \left(x\right) \ge 0$ for $1 \le x \le 3$ with the exclusion of $x = 2$ where it is undefined.

Nov 30, 2016

The answer is $x \in \left[1 , 2 \left[\cup\right] 2 , 3\right]$

#### Explanation:

Let $f \left(x\right) = \frac{\left(x - 1\right) \left(3 - x\right)}{x - 2} ^ 2$

The domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R} - \left\{2\right\}$

Also the denominator is $> 0$, $\forall x \in {D}_{f} \left(x\right)$

So, we make the sign chart

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$1$$\textcolor{w h i t e}{a a a a}$$3$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x - 1$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$3 - x$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$

$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$

So,

$f \left(x\right) \ge 0$ when $x \in \left[1 , 2 \left[\cup\right] 2 , 3\right]$

graph{((x-1)(3-x))/(x-2)^2 [-12.66, 12.65, -6.33, 6.33]}