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Math Help - Find zeroes of this function (difficult)

  1. #1
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    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 Attached Thumbnails Find zeroes of this function (difficult)-math.jpg  
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by mwok View Post
    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
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    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?
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by mwok View Post
    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
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    Quote Originally Posted by Chris L T521 View Post
    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?
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  6. #6
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by mwok View Post
    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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