Question

How do use the first derivative test to determine the local extrema `f(x) = x / (x^2+1)`?

Answer 1

See the explanation.

Explanation:

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

`f'(x)=0 <=> 1-x^2=0 <=> x=1 vv x=-1`

`f'(x)>0` `AAx in (-1,1)` function increasing
`f'(x)<0` `AAx in (-oo,-1)uu(1,oo)` function decreasing

`f'(x)` changes sign at `x=-1` and `f(x)` has a minimum value `f_min=f(-1)=-1/2`

`f'(x)` changes sign at `x=1` and `f(x)` has a maximum value `f_max=f(1)=1/2`