# Thread:

1. ## 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?
Attached Thumbnails

2. 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 $e^{2x}\neq0~\forall~x\in\mathbb{R}$

Thus, when solving for the zeros, we can rewrite $2x^2e^{2x}+2xe^{2x}+e^{2x}+2xe^{2x}=0\implies e^{2x}\left[2x^2+4x+1\right]=0$ as $2x^2+4x+1=0$

This doesn't factor nicely, so I would recommend using the quadratic formula.

--Chris

3. 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?

4. 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.

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

--Chris

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

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

--Chris

Right. What about the other factor though (e^2x)? Do I solve it with a log equation?

6. 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, $e^{2x}{\color{red}\neq}0$ for all values of $x\in\mathbb{R}$.

This can be shown by solving that equation:

$e^{2x}=0\implies 2x=\ln 0$.

However, $\ln 0$ is undefined. Thus, there is no value of x that can cause this equation to be true.

--Chris
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