# How do you solve 1/(x-2)^2<=1 using a sign chart?

Dec 17, 2016

The answer is x in ] -oo,1 ] uu [3, +oo[

#### Explanation:

We rewrite the equation as

$1 - \frac{1}{x - 2} ^ 2 \ge 0$

We do some simplifications

$\frac{{\left(x - 2\right)}^{2} - 1}{x - 2} ^ 2 \ge 0$

$\frac{\left(x - 2 - 1\right) \left(x - 2 + 1\right)}{x - 2} ^ 2 \ge 0$

$\frac{\left(x - 3\right) \left(x - 1\right)}{x - 2} ^ 2 \ge 0$

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

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

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

We can do 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}$$2$$\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 a}$$+$∣∣$\textcolor{w h i t e}{a a}$$+$$\textcolor{w h i t e}{a a a}$$+$

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

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

Therefore,

$f \left(x\right) \ge 0$, when x in ] -oo,1 ] uu [3, +oo[

graph{(y-((x-1)(x-3))/(x-2)^2)=0 [-8.89, 8.89, -4.444, 4.445]}