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Math Help - Please help me solve

  1. #1
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    Please help me solve

    List the values at which the function

    <br /> <br />
f(x) = \frac{1}{x^2+9}<br />

    Is not defined.

    This question does not have an answer in my text book so I cannot deduce where I have gone wrong, though actually I haven't gotten anywhere...just staring at it and repeating it to myself. I've tried factorising it out, but that doesn't work see (x+3)(x-3)..... {x^2} must be a positive number right?
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  2. #2
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    well you did the right thing in factorising the bottom line. The function is not defined if the denominator is 0

    => f(x) not defined at  x^{2} + 9 = 0

    => not defined  x = 3 and  x = -3
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  3. #3
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by Tekken View Post
    well you did the right thing in factorising the bottom line. The function is not defined if the denominator is 0

    => f(x) not defined at  x^{2} + 9 = 0

    => not defined  x = 3 and  x = -3
    f(3) = \frac{1}{3^2+9} = \frac{1}{18} \text{ and } f(-3) = \frac{1}{(-3)^2+9} = \frac{1}{18} which is clearly defined.


    (x+3)(x-3) = x^2-9. Note that (-3)^2 = 3^2 = 9 and hence x^2+9 \geq 9 over the real numbers.

    This means there are no real values of x for which f(x) is undefined.


    There are complex ones though: 3i and it's conjugate -3i. This comes from solving x^2+9 =0 over the complex numbers.


    Rewrite f(x) in terms of i (where i^2 = -1)

    x^2 + 9 = x^2 - (3i)^2

    The difference of two squares can now be used

    x^2-(3i)^2 = (x-3i)(x+3i) = 0
    Last edited by e^(i*pi); May 16th 2010 at 04:31 AM. Reason: oops missed a > sign
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  4. #4
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    Ah so a trick question. I always fall for these. I am so mathematically challenged

    Thank you for your help
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