differentiate f(x) = 2xe^x

answer I got was 2xe^x + 2e^x = 2e^x(x+1) equal to zero and critical values x=-1

2. Originally Posted by weezie23
differentiate f(x) = 2xe^x

answer I got was 2xe^x + 2e^x = 2e^x(x+1) equal to zero and critical values x=-1
Correct!

--Chris