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?
Note that $\displaystyle e^{2x}\neq0~\forall~x\in\mathbb{R}$
Thus, when solving for the zeros, we can rewrite $\displaystyle 2x^2e^{2x}+2xe^{2x}+e^{2x}+2xe^{2x}=0\implies e^{2x}\left[2x^2+4x+1\right]=0$ as $\displaystyle 2x^2+4x+1=0$
This doesn't factor nicely, so I would recommend using the quadratic formula.
--Chris
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?
As I mentioned earlier, $\displaystyle e^{2x}{\color{red}\neq}0$ for all values of $\displaystyle x\in\mathbb{R}$.
This can be shown by solving that equation:
$\displaystyle e^{2x}=0\implies 2x=\ln 0$.
However, $\displaystyle \ln 0$ is undefined. Thus, there is no value of x that can cause this equation to be true.
--Chris