Question

Is `f(x)=xe^x-x^2e^x` increasing or decreasing at `x=0`?

Answer 1

`f(x)=xe^x-x^2e^x` is increasing at `x=0`.

Explanation:

To find whether a function is increasing or decreasing at a point, we should find the value of its derivative at that point. If the derivative is positive, it is increasing and if it is negative, it is decreasing.

As `f(x)=xe^x-x^2e^x`

`(df)/(dx)=1*e^x+x*e^x-(2xe^x+x^2e^x)` or

`(df)/(dx)=e^x+xe^x-2xe^x-x^2e^x=e^x(1-x-x^2)`

and at `x=0`, `(df)/(dx)=e^0(1-0-0^2)=1xx1=1`

as it is positive, `f(x)=xe^x-x^2e^x` is increasing at `x=0`.

graph{xe^x-x^2e^x [-5, 5, -2.5, 2.5]}