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Thread: 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 $\displaystyle g^2(x)$ 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 $\displaystyle g^2+ 1= (3x+1)g$, 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
    $\displaystyle f(x) = \frac{x+1}{x}$
    If it is, then you're right: $\displaystyle g(x) + 1 = g(x)(3x+1)$. So you can now say:
    $\displaystyle g(x) + 1 = 3x.g(x)+ g(x)$

    $\displaystyle \Rightarrow 3x.g(x) =1$

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

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

    $\displaystyle =1+3x$
    If $\displaystyle f(x) = x + \frac1x$, 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, $\displaystyle f(x)= x+ 1/x= x+ \frac{1}{x}$, is not the same as $\displaystyle (x+1)/x= \frac{x+ 1}{x}$.

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