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

Jan 5, 2016

Increasing.

#### Explanation:

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

To find $f ' \left(x\right)$, use the quotient rule.

$f ' \left(x\right) = \frac{\left(2 x - 4\right) \frac{d}{\mathrm{dx}} \left(1 - {e}^{2 x}\right) - \left(1 - {e}^{2 x}\right) \frac{d}{\mathrm{dx}} \left(2 x - 4\right)}{2 x - 4} ^ 2$

The derivative of each part:

The first requires the chain rule:$\frac{d}{\mathrm{dx}} \left({e}^{u}\right) = {e}^{u} \cdot u '$

$\frac{d}{\mathrm{dx}} \left(1 - {e}^{2 x}\right) = - 2 {e}^{2 x}$

$\frac{d}{\mathrm{dx}} \left(2 x - 4\right) = 2$

Thus,

$f ' \left(x\right) = \frac{\left(2 x - 4\right) \left(- 2 {e}^{2 x}\right) - 2 \left(1 - {e}^{2 x}\right)}{2 x - 4} ^ 2$

$f ' \left(x\right) = \frac{- 4 x {e}^{2 x} + 10 {e}^{2 x} + 2}{2 x - 4} ^ 2$

Find $f ' \left(- 1\right)$ 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.

$f ' \left(- 1\right) = \frac{- 4 \left(- 2\right) {e}^{-} 2 + 10 {e}^{-} 2 + 2}{- 2 - 4} ^ 2 = \frac{\frac{18}{e} ^ 2 + 2}{36}$

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]}