# Is f(x)=1/e^x increasing or decreasing at x=0?

Feb 6, 2016

Decreasing

#### Explanation:

First, recognize that $f \left(x\right)$ can be written as

$f \left(x\right) = {e}^{-} x$

To determine whether this is increasing or decreasing at a point, we use the sign of the first derivative.

• If $f ' \left(0\right) < 0$, then $f \left(x\right)$ is decreasing at $x = 0$.
• If $f ' \left(0\right) > 0$, then $f \left(x\right)$ is increasing at $x = 0$.

Now, to find the derivative, we will use the chain rule. In the case of an exponential function with base $e$, the chain rule states that

$\frac{d}{\mathrm{dx}} \left({e}^{u}\right) = {e}^{u} \cdot u '$

Here, $u = - x$, so

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

Find the sign of the derivative at $x = 0$:

$f ' \left(0\right) = - {e}^{-} 0 = - {e}^{0} = - 1$

Recall that anything (other than $0$) to the $0$ power is $1$.

Since $- 1 < 0$, the function is decreasing at $x = 0$.

We can check a graph of the original function:

graph{e^-x [-10, 15.31, -4.05, 8.6]}