# Is f(x)=1/x-1/x^3+1/x^5 increasing or decreasing at x=2?

Jan 10, 2016

Decreasing.

#### Explanation:

$f ' \left(x\right) = - \frac{1}{x} ^ 2 + \frac{3}{x} ^ 4 - \frac{5}{x} ^ 6$

$f ' \left(2\right) = - \frac{9}{64}$

Decreasing at $x = 2$.

To see why, consider a small change in $x$, $\delta x$, near the neighborhood of $x = 2$.

We can approximate $f \left(2 + \delta x\right)$ as

$f \left(2\right) + f ' \left(2\right) \delta x$.

This approximation is most accurate for small values of $\delta x$.

Since $f ' \left(2\right)$ is negative,

for sufficiently small values of $\delta x > 0$, $f \left(2 + \delta x\right) < f \left(2\right)$,

and for sufficiently small values of $\delta x < 0$, $f \left(2 + \delta x\right) > f \left(2\right)$.

Therefore, $f \left(x\right)$ is decreasing at $x = 2$.