Question

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

Answer 1

The intervals of concavity is `(-oo,3)`
The intervals of convexity is `(3,+oo)`

Explanation:

We calculate the second derivative

`y=1/(x-3)`

`y'=-1/(x-3)^2`

`y''=2/(x-3)^3`

`y''=0` when `2/(x-3)^3=0`

There is no point of inflexion

We construct a chart

`color(white)(aaaa)` `Interval` `color(white)(aaaa)` `(-oo,3)` `color(white)(aaaa)` `(3,+oo)`

`color(white)(aaaa)` `"sign"` `color(white)(a)` `y''` `color(white)(aaaaaaaaa)` `-` `color(white)(aaaaaaaa)` `+`

`color(white)(aaaa)` `` `color(white)(aaa)` `y` `color(white)(aaaaaaaaaaaa)` `nn` `color(white)(aaaaaaaa)` `uu`

Answer 2

We do not have a point of inflection. The function is concave in the interval `(-oo,3)` and is convex in the interval if `(3,oo)`

Explanation:

Points of inflection appear at points where `(d^2y)/(dx^2)=0`

Here `y=1/(x-3)` and as such `(dy)/(dx)=-1/(x-3)^2`

Therefore `(dy)/(dx)` is always negative i.e. function is always declining

and as `(d^2y)/(dx^2)=-(-2)/(x-3)^3=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''(x)<0` in an interval and is convex in an interval if `f''(x)>0`.

Here, function is concave in the interval `(-oo,3)` and is convex in the interval if `(3,oo)`.

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