Find the zeros of function
f(x) = x^2(2e^2x) + 2xe^2x + e^2x + 2xe^2x)
Attached is what I got. Is it correct?
Ahh, you have to take the commons out and then use the quad formula. Nice!
Question...I used the quad formula and ended up with
x = (-2 +/- sqrt(2)) / 2
Do I make it equal zero and then solve it further OR is that the direct zero?
Also, what about the other factor e^2x....how would you solve that for zero?
e^2x = 0
I'm guessing you have to use the log equation on that?