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

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Answer 1 · 2016-01-10T07:56:13

Decreasing.

#f'(x) = -1/x^2 + 3/x^4 - 5/x^6#

#f'(2) = -9/64#

Decreasing at #x = 2#.

To see why, consider a small change in #x#, #deltax#, near the neighborhood of #x=2#.

We can approximate #f(2+deltax)# as

#f(2) + f'(2)deltax#.

This approximation is most accurate for small values of #deltax#.

Since #f'(2)# is negative,

for sufficiently small values of #deltax > 0#, #f(2+deltax)<f(2)#,

and for sufficiently small values of #deltax < 0#, #f(2+deltax)>f(2)#.

Therefore, #f(x)# is decreasing at #x=2#.