Question

What is the instantaneous rate of change of `f(x)=xe^x-x^2e^x` at `x=-1`?

Answer 1

`1/e`

Explanation:

The instantaneous rate of change of a function `f(x)` at a given point is the slope of the tangent line at that point, that is, the derivative `f'(x)` evaluated at that point.

In this case, we have

`f'(x) = xe^x+e^x-x^2e^x-2xe^x`

`=-e^x(x^2+x-1)`

Then, evaluated at `x=-1` we have

`f'(-1) = -e^(-1)((-1)^2+(-1)-1)`

`=-1/e(-1)`

`=1/e`