Question

How do you find the x coordinates of all points of inflection, final all discontinuities, and find the open intervals of concavity for `f(x)=(x^2+1)/(x^2-4)`?

Answer 1

There are no points of inflection.
The discontinuities are when `x=-2` and `x=2`
`f(x)` is concave up for `x in ]-oo,-2[ uu]2.+oo[`
`f(x)` is concave down for `x in ]-2,2[`

Explanation:

As we cannot divide by `0`, `x!=-2` and `x!=2`

The domain of `f(x)` is `d_f(x)=RR-{-2,2}`

Therefore, the discontinuities are when `x=-2` and `x=2`

The derivative of a quotient is

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

We calculate the first derivative of `f(x)`

`u=x^2+1`, `=>`, `u'=2x`

`v=x^2-4`, `=>`, `v'=2x`

`f'(x)=(2x(x^2-4)-2x(x^2+1))/(x^2-4)^2`

`=(2x(x^2-4-x^2-1))/(x^2-4)^2`

`=(-10x)/(x^2-4)^2`

We can calculate the second derivative of `f(x)`

`u=-10x`, `=>`, `u'=-10`

`v=(x^2-4)^2`, `=>`, `v'=2(x^2-4)*2x=4x(x^2-4)`

`f''(x)=(-10(x^2-4)^2+10x(4x(x^2-4)))/(x^2-4)^4`

`=((x^2-4)(-10x^2+40+40x^2))/(x^2-4)^4`

`=(30x^2+40)/(x^2-4)^3`

`=(10(3x^2+4))/(x^2-4)^3`

The points of inflexion are when, `f''(x)=0`

`f''(x)!=0`, there are no points of inflection.

Let build the chart for concavities

`color(white)(aaaa)` `Intervals` `color(white)(aaaa)` `]-oo,-2[` `color(white)(aaaa)` `]-2,2[` `color(white)(aaaa)` `]2,+oo[`

`color(white)(aaaa)` `x+2` `color(white)(aaaaaaaaaaa)` `-` `color(white)(aaaaaaaaaaa)` `+` `color(white)(aaaaaaa)` `+`

`color(white)(aaaa)` `x-2` `color(white)(aaaaaaaaaaa)` `-` `color(white)(aaaaaaaaaaa)` `-` `color(white)(aaaaaaa)` `+`

`color(white)(aaaa)` `Sign f''(x)` `color(white)(aaaaaaa)` `+` `color(white)(aaaaaaaaaaa)` `-` `color(white)(aaaaaaa)` `+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaaaaaaaa)` `uu` `color(white)(aaaaaaaaaaa)` `nn` `color(white)(aaaaaaaa)` `uu`

Therefore,

`f(x)` is concave up for `x in ]-oo,-2[ uu]2.+oo[`

`f(x)` is concave down for `x in ]-2,2[`