Question

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

Answer 1

For `x=1` the function `f(x)` has a local maximum.

Explanation:

Calculate the first derivative of the function:

`f(x) = x/e^x = xe^(-x)`

`(df)/dx = e^(-x)-xe^(-x) = e^(-x)(1-x) =(1-x)/e^x`

We can therefore see that `x=1` is a stationary point for `f(x)`, so the function is neither increasing nor decreasing but has a local extremum.

Looking at `(df)/dx` around `x=1` we can see that:

`(df)/dx > 0` for `x < 1`

`(df)/dx < 0` for `x > 1`

which means that `f(x)` is increasing from `-oo` to `x=1` and decreasing from `x=1` to `+oo`, so for `x=1` it has a local maximum.

graph{x/e^x [-2.23, 7.77, -3.42, 1.58]}