Question

What is the instantaneous rate of change of `f(x)=xe^x-(x+2)e^(x-1)` at `x=0`?

Answer 1

I got: `(e-3)/e=-0.10364`

Explanation:

To evaluate the instantaneous rate of change we need to evaluate the Derivative of our function and evaluate it at `x=0`.
So we get:
`f'(x)=e^x+xe^x-e^(x-1)-(x+2)e^(x-1)`

let us evaluate it at `x=0`
`f'(0)=e^0+0e^0-e^(0-1)-(0+2)e^(0-1)=1-1/e-2/e=(e-3)/e=-0.10364`