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

1 Answer
Aug 21, 2017

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!