# If  f(x)= (x^2)/(x-2)^2, what are the points of inflection, concavity and critical points?

Jun 16, 2017

The critical point is $= \left(0 , 0\right)$. The inflection point is $= \left(- 1 , \frac{1}{9}\right)$. See below for the concavities.

#### Explanation:

We need

$\left(\frac{u}{v}\right) ' = \frac{u ' v - u v '}{{v}^{2}}$

Let's calculate the first derivative

$f \left(x\right) = {x}^{2} / {\left(x - 2\right)}^{2}$

$u = {x}^{2}$, $\implies$, $u ' = 2 x$

$v = {\left(x - 2\right)}^{2}$, $\implies$, $v ' = 2 \left(x - 2\right)$

$f ' \left(x\right) = \frac{2 x \cdot {\left(x - 2\right)}^{2} - {x}^{2} \cdot \left(2 \left(x - 2\right)\right)}{x - 2} ^ 4$

$= \frac{\left(x - 2\right) \left(2 x \left(x - 2\right) - 2 {x}^{2}\right)}{x - 2} ^ 4$

$= \frac{2 {x}^{2} - 4 x - 2 {x}^{2}}{x - 2} ^ 3$

$= \frac{- 4 x}{x - 2} ^ 3$

$f ' \left(x\right) = 0$, when $x = 0$

We build our first 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 a a a a}$$0$$\textcolor{w h i t e}{a a a a a a a a a a}$$2$$\textcolor{w h i t e}{a a a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$- 4 x$$\textcolor{w h i t e}{a a a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$0$$\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 a}$$-$

$\textcolor{w h i t e}{a a a a}$${\left(x - 2\right)}^{3}$$\textcolor{w h i t e}{a a a}$$-$$\textcolor{w h i t e}{a a a a}$$0$$\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 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 a}$$0$$\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 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 a}$↘$\textcolor{w h i t e}{a a a}$$0$$\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}$↘

The critical point is when $x = 0$

The intervals of increasing are $x \in \left(0 , 2\right)$ and decreasing are $x \in \left(- \infty , 0\right) \cup \left(2 , + \infty\right)$

Now, we calculate the second derivative

$u = - 4 x$, $\implies$, $u ' = - 4$

$v = {\left(x - 2\right)}^{3}$, $\implies$, $v ' = 3 {\left(x - 2\right)}^{2}$

So,

$f ' ' \left(x\right) = \frac{- 4 {\left(x - 2\right)}^{3} + 4 x \cdot 3 {\left(x - 2\right)}^{2}}{x - 2} ^ 6$

$= \frac{- 4 {\left(x - 2\right)}^{2} \left(\left(x - 2\right) - 3 x\right)}{x - 2} ^ 6$

$= - 4 \frac{x - 2 - 3 x}{x - 2} ^ 4$

$= - 4 \frac{- 2 x - 2}{x - 2} ^ 4$

$= \frac{8 \left(x + 1\right)}{x - 2} ^ 4$

Therefore,

$f ' ' \left(x\right) = 0$ when $x = - 1$, this is the inflexion point

The chart is as follows

$\textcolor{w h i t e}{a a a a}$$I$$\textcolor{w h i t e}{a a a a a a}$$\left(- \infty , - 1\right)$$\textcolor{w h i t e}{a a a a}$$\left(- 1 , 2\right)$$\textcolor{w h i t e}{a a a a}$$\left(2 , + \infty\right)$

$\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 a a}$$-$$\textcolor{w h i t e}{a a a a a a a a a}$$+$$\textcolor{w h i t e}{a a a a 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 a a}$$\cap$$\textcolor{w h i t e}{a a a a a a a a a}$$\cup$$\textcolor{w h i t e}{a a a a a a a a}$$\cup$

The curve is concave is up when $x \in \left(- 1 , 2\right) \cup \left(2 , + \infty\right)$

The curve is concave down when $x \in \left(- \infty , - 1\right)$

Also,

$f ' ' \left(0\right) > 0$, this is a minimum

graph{x^2/(x-2)^2 [-11.36, 11.14, -1.58, 9.67]}