Is #f(x)=x/(2-e^x)# increasing or decreasing at #x=0#?

1 Answer
Jun 16, 2016

#f(x)=x/(2-e^x)# is increasing at #x-0#.

Explanation:

If a function is increasing at a point, value of its derivative at that point is positive and if function is decreasing at the point, value of its derivative is negative at that point.

As #f(x)=x/(2-e^x)#

#(df)/(dx)=((2-e^x)xx1-x xx(-e^x))/(2-e^x)^2#

= #(2-e^x+xe^x)/(2-e^x)^2#

At #x=0#, #(df)/(dx)=(2-e^0+0e^0)/(2-e^0)^2#

= #(2-1+0)/(2-1)^2=1/1=1#

Hence the function #f(x)=x/(2-e^x)# is increasing at #x-0#.

graph{x/(2-e^x) [-2.5, 2.5, -1.25, 1.25]}