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E^(x^2) = 0

  1. Nov 18, 2011 #1
    1. The problem statement, all variables and given/known data

    e^(x^2) = 0

    2. Relevant equations

    How do you solve this for x? Never had to solve a power to a power before.

    3. The attempt at a solution



    e^x^2 = 0

    lne^x^2 = ln0

    log(base e)of e^(x^2) = 1

    e^(1) = e^(x^2)

    1 = x^(2)

    x = √1

    If this is the correct answer, is it the only answer?
     
  2. jcsd
  3. Nov 18, 2011 #2

    D H

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    That is not the correct answer.
     
  4. Nov 18, 2011 #3

    gb7nash

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    Think about it. Can a positive constant raised to any number equal to 0?
     
  5. Nov 18, 2011 #4
    Think about what gb7 said, also just so you know ln(0) is undefined, it's not equal to 0!
     
  6. Nov 18, 2011 #5
    Ohh!! so there must be no solution then, right?

    And double Ohh!!, ln0 is undefined...its ln1 that equals zero!!

    Is my assumption of no solution correct?
     
  7. Nov 18, 2011 #6
    That is correct :D
     
  8. Nov 18, 2011 #7
    Awsome!! thanks.

    So if f(x) = (2x^(2) + 1)(e^(x^2))

    and f(x) = 0

    so that (2x^(2) + 1)(e^(x^2)) = 0 and I solve for x,

    e^(x^2) = 0 is no solution

    and (2x^(2) + 1) = 0 is:

    x^2 = -1/2 so

    x = √-1/2

    And since √-1/2 has a negative on the inside, it's a complex number and has no solution also, right?

    Therefore, are there no real roots for (2x^(2) + 1)(e^(x^2)) = 0 ?

    And since this happens to be a first derivative function, does that then mean there are no critical points on the original function xe^(x^2)?
     
  9. Nov 18, 2011 #8

    eumyang

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    That's right.
     
  10. Nov 19, 2011 #9

    dextercioby

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    Yes, the function's increasing as x takes all real values from -infinity to +infinity.
     
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