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

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.

$f \left(x\right) = {x}^{-} 1 - {x}^{-} 3 + {x}^{5} - x$

$f ' \left(x\right) = - {x}^{-} 2 - \left(- 3 {x}^{-} 4\right) + 5 {x}^{4} - 1$

Multiplying by ${x}^{4} / {x}^{4}$:

$f ' \left(x\right) = \frac{- {x}^{2} + 3 + 5 {x}^{8} - {x}^{4}}{x} ^ 4 = \frac{5 {x}^{8} - {x}^{4} - {x}^{2} + 3}{x} ^ 4$

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

$f ' \left(x\right)$:

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!