How do you find the critical numbers for #f(x)= (x-1) e^x# to determine the maximum and minimum?

1 Answer
Feb 28, 2017

Answer:

The only critical point is for #x=0# and the function has a local minimum in that point.

Explanation:

Critical points are found by equating the first derivative to zero:

#f'(x) = d/dx ((x-1)e^x) = d/dx (x-1)e^x + (x-1)d/dx e^x = e^x +(x-1)e^x =xe^x#

Solving:

#xe^x = 0#

the only critical point is #x=0#

Now consider that:

#f'(x) = xe^x < 0# for #x<0#

#f'(x) = xe^x > 0# for #x>0#

We have then that #f(x)# is decreasing in #(-oo,0)# and increasing in #(0,+oo)#, therefore #x=0# is a local minimum.

graph{(x-1)e^x [-10, 10, -5, 5]}