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

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Answer 1 · 2017-02-04T15:15:51

Decreasing.

Rewrite

#f(x) = x^-1 - x^-3 + x^-5#

Differentiate:

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

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

Let #x = a# be your point. If #f'(a) > 0#, then #f(x)# is increasing at #a#. If #f'(a) < 0#, then #f(x)# is decreasing at #a#.

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

Hence, #f(x)# is decreasing when #x = -1#.

Hopefully this helps!