Question

What are the points of inflection of #f(x)=x^2 / (x^2 + 49) #?

Answer 1

The points of inflection are `(7/sqrt3,1/4)` and `(-7/sqrt3,1/4)`

Explanation:

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

Let start by calculating the derivatves, we use `f'(u/v)=(u'v-uv')/v^2`

So `f'(x)=(2x(x^2+49)-x^2(2x))/(x^2+49)^2=(98x)/(x^2+49)^2`

And
`f''(x)=(98(x^2+49)^2-98x(x^2+49)*2*2x)/((x^2+49)^4)`

`=(98(x^2+49)((x^2+49)-4x^2))/((x^2+49)^4)`
`=(98(49-3x^2))/((x^2+49)^3`
so `f''(x)=0` when `49-3x^2=0`

that is `x=+-7/sqrt3`
And `y=(49/3)/(49/3+49)=1/4`
So the points are `(7/sqrt3,1/4)` and `(-7/sqrt3,1/4)`