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

Question

1564 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-19T23:31:06

decreasing.

If the function is increasing, #f'(x) > 0#. If the function is decreasing, #f'(x) < 0#.

#f'(x) = (1(x + x^2) - x(1 + 2x))/(x + x^2)^2 - 1/x#

We evaluate #x = 1# within the derivative.

#f'(1) = (1(1 + 1^2) - 1(1 + 2(1)))/(1 + 1^2)^2#

#f'(1) = (2 - 3)/(2)^2#

#f'(1) = -1/4#

#-1/4 < 0#, so #f(x)# is decreasing at #x= 1#.

Hopefully this helps!