Question

How do you find points of inflection and determine the intervals of concavity given `y=(3x+2)/(x-2)`?

Answer 1

There are no points of inflection.
The interval of concavity is `(-oo,2)` and the interval of convexity is `(2,+oo)`

Explanation:

Calculate the first derivative

`y=(3x+2)/(x-2)`

This is a quotient `u/v` and the derivative is

`(u/v)'=(u'v-uv')/(v^2)`

Here,

`u=3x+2`, `=>`, `u'=3`

`v=x-2`, `=>`, `v'=1`

Therefore,

`dy/dx=(3(x-2)-1(3x+2))/(x-2)^2=(3x-6-3x-2)/(x-2)^2`

`=-8/(x-2)^2`

And the second derivative is

`(d^2y)/dx^2=(-8(x-2)^-2)'=-8*-2/(x-2)^3=16/(x-2)^3`

The points of inflection are when `(d^2y)/dx^2=0`

Here, the second derivative is `!=0`

So, there are no points of inflection

Build a sign chart for the concavity

`color(white)(aaaa)` `Interval` `color(white)(aaaa)` `(-oo,2)` `color(white)(aaaa)` `(2,+oo)`

`color(white)(aaaa)` `Sign (d^2y)/dx^2` `color(white)(aaaaaaa)` `-` `color(white)(aaaaaaaa)` `+`

`color(white)(aaaaaaa)` `y` `color(white)(aaaaaaaaaaa)` `nn` `color(white)(aaaaaaaa)` `uu`

The interval of concavity is `(-oo,2)` and the interval of convexity is `(2,+oo)`

graph{(3x+2)/(x-2) [-41.1, 41.08, -20.56, 20.56]}