# Find zeroes of this function (difficult)

• Oct 30th 2008, 09:53 AM
mwok
Find zeroes of this function (difficult)
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?
• Oct 30th 2008, 10:01 AM
Chris L T521
Quote:

Originally Posted by mwok
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
• Oct 30th 2008, 10:15 AM
mwok
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?
• Oct 30th 2008, 10:18 AM
Chris L T521
Quote:

Originally Posted by mwok
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?

Those are the direct zeros. :D

Just keep in mind that there are two of them: $\displaystyle x=\tfrac{1}{2}\left[-2+\sqrt{2}\right]$ or $\displaystyle x=-\tfrac{1}{2}\left[2+\sqrt{2}\right]$

--Chris
• Oct 30th 2008, 10:20 AM
mwok
Quote:

Originally Posted by Chris L T521
Those are the direct zeros. :D

Just keep in mind that there are two of them: $\displaystyle x=\tfrac{1}{2}\left[-2+\sqrt{2}\right]$ or $\displaystyle x=-\tfrac{1}{2}\left[2+\sqrt{2}\right]$

--Chris

Right. What about the other factor though (e^2x)? Do I solve it with a log equation?
• Oct 30th 2008, 10:24 AM
Chris L T521
Quote:

Originally Posted by mwok
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