Question

Is `f(x)=x^2e^x` increasing or decreasing at `x=1`?

Answer 1

To know that, you are finding the slope at `x = 1`, and seeing if it is positive or negative (or `0`, but in that case it would be an extremum).

`(df)/(dx) = d/(dx)[x^2e^x]`

`= x^2e^x + 2xe^x` (product rule)

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

Thus, at `x = 1`:

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

`= 3e > 0`

So, the function is increasing at `x = 1`.