Question

What are the local extrema of `f(x)= x^2/(x^2-3x-5)`?

Answer 1

`MAX(0;0)` and `MIN(-10/3,20/29)`

Explanation:

We compute
`f'(x)=-x(3x+10)/(x^2-3x-5)^2`
`f''(x)=2(3x^2+15x^2+25)/(x^2-3x-5)^3`
so
`f'(x)=0`
if
`x=0` or
`x=-10/3`
we have further
`f''(0)=-2/5<0`
and
`f''(-10/3)=162/4205>0`