Question

What are the local extrema of `f(x)= 1/x-1/x^3+x^5-x`?

Answer 1

There are no local extrema.

Explanation:

Local extrema could occur when `f'=0` and when `f'` switches from positive to negative or vice versa.

`f(x)=x^-1-x^-3+x^5-x`

`f'(x)=-x^-2-(-3x^-4)+5x^4-1`

Multiplying by `x^4/x^4`:

`f'(x)=(-x^2+3+5x^8-x^4)/x^4=(5x^8-x^4-x^2+3)/x^4`

Local extrema could occur when `f'=0`. Since we can't solve for when this happens algebraically, let's graph `f'`:

`f'(x)`:

graph{(5x^8-x^4-x^2+3)/x^4 [-5, 5, -10.93, 55]}

`f'` has no zeros. Thus, `f` has no extrema.

We can check with a graph of `f`:

graph{x^-1-x^-3+x^5-x [-5, 5, -118.6, 152.4]}

No extrema!