# Is f(x)=(x-e^x)/(x-1)^3 increasing or decreasing at x=2?

Sep 29, 2017

Increasing.

#### Explanation:

To know if a function $f \left(x\right)$ is increasing or decreasing at some point, you must find the derivative $f ' \left(x\right)$ of the function and check the sign of the value of the derivative. If $f ' \left(x\right) > 0$, the function is said to be increasing. If $f ' \left(x\right) < 0$, the function is said to be decreasing.

Begin by using the Quotient Rule to find the derivative $f ' \left(x\right)$:

$f ' \left(x\right) = \frac{{\left(x - 1\right)}^{3} \cdot \left(1 - {e}^{x}\right) - \left(x - {e}^{x}\right) \cdot \left(3 \cdot {\left(x - 1\right)}^{2}\right)}{{\left(x - 1\right)}^{3}} ^ 2$
$= \frac{{\left(x - 1\right)}^{3} \cdot \left(1 - {e}^{x}\right) - 3 \left(x - {e}^{x}\right) {\left(x - 1\right)}^{2}}{x - 1} ^ 6$

Now we can evaluate $f ' \left(2\right)$:

$f ' \left(2\right) = \frac{{\left(2 - 1\right)}^{3} \cdot \left(1 - {e}^{2}\right) - 3 \left(2 - {e}^{2}\right) {\left(2 - 1\right)}^{2}}{2 - 1} ^ 6$
$= \frac{\left(1 - {e}^{2}\right) - 3 \left(2 - {e}^{2}\right)}{1} = \left(1 - {e}^{2} - 6 + 3 {e}^{2}\right)$
$= 2 {e}^{2} - 5 \approx 9.778 > 0$

Since $f ' \left(2\right) > 0$, we say $f \left(x\right)$ is increasing at $x = 2$.

This can also be verified by looking at the graph as a support for our work:

graph{(x-e^x)/(x-1)^3 [-5.484, 12.296, -7.46, 1.425]}