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Math Help - f(g(x)) functions

  1. #1
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    f(g(x)) functions

    f(x) = 1/(2x)
    g(x) = (3x-1)^1/2

    I got the domains of f(x) and g(x) and f(g(x)). I even got the value of g(f(x))=3/2. My algebra falters with the problem f(g(x)):

    1
    ---------
    2((3x-1)^1/2)


    I'm not sure where to start after setting it equal to 0.

    edit: Sorry for being vague. I need to solve for zeros.
    Last edited by bloodninja; October 18th 2009 at 11:22 AM.
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  2. #2
    Senior Member MacstersUndead's Avatar
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    First Note: f(g(x)) or g(f(x)) cannot be a value, since x isn't defined.

    Second Note:
    Consider that: x^1/2 = sqrt(x) and that the restriction of the domain for any radical function is that whatever is under the square root must be greater than or equal to 0.
    EDIT:// if a single radical term is in the denominator, then it is strictly greater than 0. If there instead additional terms in the denominator, the entire denominator cannot equal 0.

    If you wanted to set f(g(x)) = 0, then you will get the zeroes of the function, and since you know for square roots, the function tends to the right of the real line, then you can use it. use order of operations to solve. I prefer using the above method instead for more complex situations.
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  3. #3
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by bloodninja View Post
    f(x) = 1/2x
    g(x) = (3x-1)^1/2

    I got the domains of f(x) and g(x) and f(g(x)). I even got the value of g(f(x))=3/2. My algebra falters with the problem f(g(x)):

    1
    ---------
    2((3x-1)^1/2)


    I'm not sure where to start after setting it equal to 0.
    Does g(x) = \frac{1}{2x} or \frac{1}{2}x

    If the former: f[g(x)] = \frac{1}{2\sqrt{3x-1}}

    If the latter: \frac{1}{2}\sqrt{3x-1}
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  4. #4
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    Thanks guys, I should have specified that I need to find the zeros. I know what the order of operations are, but uh.. am retarded otherwise. I need help actually solving it.
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  5. #5
    Senior Member MacstersUndead's Avatar
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    Alright, I'll assume that you know BEDMAS and work from there, but you first have to clarify e^(i*pi)'s concern above.
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  6. #6
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    Sorry e^(i*pi), f(x)= 1/(2x)

    this is how I solved f(g(x))
    1
    -------------- = 0
    2(3x-1)^(1/2)


    (3x-1)^-(1/2)
    -------------- = 0
    2

    Then I multiply 2 to both sides of the equation, essentially getting rid of the 2.

    (3x-1)^(-1/2) = 0

    I can raise the entire equation to the power of -2 as well. I'm not sure if this is the correct way.

    (3x-1) = 0

    3x=1

    x=1/3

    It doesn't seem right. Plugging 1/3 back in the original equation gets me
    1
    ----------------
    2(3(1/3)-1)^1/2

    1
    -----------
    2(0)^1/2

    UNDEFINED. Maybe I'm determined that there IS a real solution, when in fact, there is none? I'm sure that my math is incorrect
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  7. #7
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    f(g(x)) = \frac{1}{2\sqrt{3x-1}}

    Find x such that f(g(x)) = 0 \Rightarrow \frac{1}{2\sqrt{3x-1}} = 0 multiply by 2\sqrt{3x-1} and get 1=0 \Rightarrow there is no solution.
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