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

Mar 6, 2016

Increasing. See explanation below.

#### Explanation:

In order to find if a function is increasing or decreasing, we take its derivative and evaluate it at the $x$-value in question. If the derivative is positive at that point, the function is increasing; if the derivative is negative, the function is decreasing.

Step 1: Find the Derivative
Since we have $x - 2$ divided by ${e}^{x}$, we need to use the quotient rule to take the derivative, which states:
$\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{u ' v - u v '}{v} ^ 2$
Where $u$ and $v$ are functions of $x$.

In our case, we have $u = x - 2$ and $v = {e}^{x}$. Taking the derivative of these two:
$u ' = 1$
$v ' = {e}^{x}$

Now we can substitute into the quotient rule:
$\frac{d}{\mathrm{dx}} \left(\frac{x - 2}{e} ^ x\right) = \frac{\left(x - 2\right) ' \left({e}^{x}\right) - \left(x - 2\right) \left({e}^{x}\right) '}{{e}^{x}} ^ 2$
= (1(e^x)-(x-2)(e^x))/(e^(2x)

We can do a little simplifying here:
= (e^x(1-(x-2)))/(e^(2x)
$= \frac{1 - x + 2}{{e}^{x}}$
$= \frac{- x + 3}{{e}^{x}}$

Step 2: Evaluate
We are being asked to find if the function is increasing or decreasing at $x = - 2$; that means we evaluate the derivative at $x = - 2$:
$= \frac{- \left(- 2\right) + 3}{{e}^{- 2}}$
$= \frac{5}{{e}^{- 2}} = 5 {e}^{2}$

We don't even need to find $5 {e}^{2}$, because it is definitely positive. And because it's positive, we can say that the function $\frac{x - 2}{e} ^ x$ is increasing at $x = - 2$. To confirm, take a look at the graph of $\frac{x - 2}{e} ^ x$ and you will see it increasing at $x = - 2$.
graph{(x-2)/e^x [-10, 10, -5, 5]}