Question

Is `f(x)=(1-e^(2x))/(2x-4)` increasing or decreasing at `x=-1`?

Answer 1

Increasing.

Explanation:

Find the sign of the derivative at `x=-1`.

To find `f'(x)`, use the quotient rule.

`f'(x)=((2x-4)d/dx(1-e^(2x))-(1-e^(2x))d/dx(2x-4))/(2x-4)^2`

The derivative of each part:

The first requires the chain rule:`d/dx(e^u)=e^u*u'`

`d/dx(1-e^(2x))=-2e^(2x)`

`d/dx(2x-4)=2`

Thus,

`f'(x)=((2x-4)(-2e^(2x))-2(1-e^(2x)))/(2x-4)^2`

`f'(x)=(-4xe^(2x)+10e^(2x)+2)/(2x-4)^2`

Find `f'(-1)` If it's greater than `0`, the function is increasing at that point. If it's less than `0`, the function is decreasing at that point.

`f'(-1)=(-4(-2)e^-2+10e^-2+2)/(-2-4)^2=(18/e^2+2)/36`

Since this is `>0`, the function is increasing when `x=-1`.

graph{(1-e^(2x))/(2x-4) [-5.93, 6.56, -2.596, 3.647]}