Question

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

Answer 1

Decreasing

Explanation:

First, recognize that `f(x)` can be written as

`f(x)=e^-x`

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

• If `f'(0)<0`, then `f(x)` is decreasing at `x=0`.
• If `f'(0)>0`, then `f(x)` 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

`d/dx(e^u)=e^u*u'`

Here, `u=-x`, so

`f'(x)=e^-x*d/dx(-x)=e^-x*(-1)=-e^-x`

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

`f'(0)=-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]}