# How do you find points of inflection and determine the intervals of concavity given y=1/(x-3)?

Apr 9, 2017

The intervals of concavity is $\left(- \infty , 3\right)$
The intervals of convexity is $\left(3 , + \infty\right)$

#### Explanation:

We calculate the second derivative

$y = \frac{1}{x - 3}$

$y ' = - \frac{1}{x - 3} ^ 2$

$y ' ' = \frac{2}{x - 3} ^ 3$

$y ' ' = 0$ when $\frac{2}{x - 3} ^ 3 = 0$

There is no point of inflexion

We construct a chart

$\textcolor{w h i t e}{a a a a}$$I n t e r v a l$$\textcolor{w h i t e}{a a a a}$$\left(- \infty , 3\right)$$\textcolor{w h i t e}{a a a a}$$\left(3 , + \infty\right)$

$\textcolor{w h i t e}{a a a a}$$\text{sign}$$\textcolor{w h i t e}{a}$$y ' '$$\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}$$+$

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

Apr 9, 2017

We do not have a point of inflection. The function is concave in the interval $\left(- \infty , 3\right)$ and is convex in the interval if $\left(3 , \infty\right)$

#### Explanation:

Points of inflection appear at points where $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 0$

Here $y = \frac{1}{x - 3}$ and as such $\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{x - 3} ^ 2$

Therefore $\frac{\mathrm{dy}}{\mathrm{dx}}$ is always negative i.e. function is always declining

and as $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - \frac{- 2}{x - 3} ^ 3 = \frac{2}{x - 3} ^ 3$, it is positive when $x > 3$ and when $x < 3$, it is negative, bur no where it is $0$,

hence we do not have a point of inflection.

A function is concave if $f ' ' \left(x\right) < 0$ in an interval and is convex in an interval if $f ' ' \left(x\right) > 0$.

Here, function is concave in the interval $\left(- \infty , 3\right)$ and is convex in the interval if $\left(3 , \infty\right)$.

graph{1/(x-3) [-10, 10, -5, 5]}