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Math Help - Finding g(x) given f(x) and f o g(x)?

  1. #1
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    Finding g(x) given f(x) and f o g(x)?

    How do I find g(x) if f(x) = x+1/x and f o g(x) = 3x+1?

    I've gotten as far as g(x) + 1 = g(x)(3x + 1)

    What do I have to do next?

    Thanks for the help!
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  2. #2
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    Quote Originally Posted by strigy View Post
    How do I find g(x) if f(x) = x+1/x and f o g(x) = 3x+1?

    I've gotten as far as g(x) + 1 = g(x)(3x + 1)
    That should be g^2(x) on the left.

    What do I have to do next?

    Thanks for the help!
    Think of g(x) as being a variable itself rather than a function:
    f(g)= g+ 1/g= 3x+1. Now solve for g. Multiplying both sides by g gives g^2+ 1= (3x+1)g, a quadratic equation. Use the quadratic formula to solve for all x.
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  3. #3
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    Uhm, how did it get to be g^2 on the left? Shouldn't the denominator on g+1/g have cancelled out when I multiplied the whole equation with g?
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  4. #4
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    Hello strigy
    Quote Originally Posted by strigy View Post
    How do I find g(x) if f(x) = x+1/x and f o g(x) = 3x+1?

    I've gotten as far as g(x) + 1 = g(x)(3x + 1)

    What do I have to do next?

    Thanks for the help!
    Having read what has been written so far, I wonder if the question is really
    f(x) = \frac{x+1}{x}
    If it is, then you're right: g(x) + 1 = g(x)(3x+1). So you can now say:
    g(x) + 1 = 3x.g(x)+ g(x)

    \Rightarrow 3x.g(x) =1

    \Rightarrow g(x) = \frac{1}{3x}
    You can now check this:
    f\circ g(x) = f\left(\frac{1}{3x}\right)
    = \frac{\dfrac{1}{3x}+1}{\dfrac{1}{3x}}

    =\frac{1+3x}{1}

    =1+3x
    If f(x) = x + \frac1x, then you'll have to do as HoI suggests and use the quadratic formula.

    Grandad
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  5. #5
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    Grandad, it is f(x) = x+1/x, thank you so much!

    HoI, thanks for showing me what to do if ever I get an equation like that.
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  6. #6
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    As you originally wrote it, f(x)= x+ 1/x= x+ \frac{1}{x}, is not the same as (x+1)/x= \frac{x+ 1}{x}.

    x+ \frac{1}{x}= \frac{x^2}{x}+ \frac{1}{x}= \frac{x^2+ 1}{x}
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  7. #7
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    Oh,ok HoI, my bad. Thanks!
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