Question

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

Answer 1

`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]}