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

1 Answer
Aug 21, 2017

Answer:

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!