First we need to find the critical point(s). The derivative is *f *'(*x*) = *xe*^{x} + *e*^{x} = *e*^{x}(*x* + 1).
This is zero only when *x* = -1. Since *e*^{x} is always positive, the sign of *f*'(*x*) is the same as the sign of ( *x* + 1 ). When *x* < -1, the quantity ( *x* + 1 ) is negative, so *f*'(*x*) is also. When *x* > -1 the quantity ( *x* + 1 ) is positive, so *f*'(*x*) is also. We therefore find this numberline:
Therefore *f* (*x*) is decreasing until *x* approaches -1, then increasing again, so at *x* = -1 there is a minimum. |