Question

How do you use the first and second derivatives to sketch `f(x) = (x+1)e^x`?

Answer 1

there is a minimum in `(-2;-e^{-2})`
and an inflection point in `(-3;-2e^{-3})`
please see the graph below

Explanation:

`f'(x)=e^x+(x+1)e^x=e^x(x+2)` whose only root is clearly in `x=-2`, besides for `x>-2` it is crescent whereas for `x<-2` it decrescent so that `x=-2` is point of minimum whose ordinate is `y=-e^{-2}`

By deriving once more we get `f''(x)=e^x+(x+2)e^x=e^x(x+3)` whose only root is for `x=-3`. So for `x> -3` the function has a positive concativity whereas for `x<-3` the concativity is negative graph{e^x*(1+x) [-4.302, 0.698, -1.02, 1.48]}