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Math Help - Functions

  1. #1
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    Functions

    I really don't get this...

    If anyone could solve the following for me and explain every step it would be great.

    If f(2x-1)=x then f(f(x))=?

    A journey towards my math enlightenment has begun
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  2. #2
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    Quote Originally Posted by ikislav View Post
    If anyone could solve the following for me and explain every step it would be great.
    If f(2x-1)=x then f(f(x))=?
    From the given f(x)=f\left(2\left(\frac{x}{2}+\frac{1}{2}\right)-1\right)=\frac{x}{2}+\frac{1}{2}.
    Now YOU finish.
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Denote t=2x-1 then, f(t)=(t+1)/2 or equivalently f(x)=(x+1)/2 .


    Edited: Sorry, I didn't see the previous post.
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  4. #4
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    Quote Originally Posted by Plato View Post
    From the given f(x)=f\left(2\left(\frac{x}{2}+\frac{1}{2}\right)-1\right)=\frac{x}{2}+\frac{1}{2}.
    Now YOU finish.
    No, I simply don't understand this...
    How did you know that f(x)=f\left(2\left(\frac{x}{2}+\frac{1}{2}\right)-1\right)=\frac{x}{2}+\frac{1}{2}?

    What you showed me has just confused me even more (if that is possible).
    Please explain to me step by step, maybe then I will figure it out.
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  5. #5
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    Quote Originally Posted by ikislav View Post
    Please explain to me step by step, maybe then I will figure it out.
    That is not how it works here.
    You should understand that this is not a tutorial service.
    Perhaps you need to hire a tutor.

    Or else you could post some of your own work on this problem or explain what you do not understand about the question.

    You are responsible for your own learning.
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ikislav View Post
    No, I simply don't understand this...
    How did you know that f(x)=f\left(2\left(\frac{x}{2}+\frac{1}{2}\right)-1\right)=\frac{x}{2}+\frac{1}{2}?

    What you showed me has just confused me even more (if that is possible).
    Please explain to me step by step, maybe then I will figure it out.
    I think FernandoRevilla's post is likely to be the most transparent.

    Let t=2x-1 then, \displaystyle x = \frac{1}{2} \cdot (t + 1). Then either follow with FernandoRevilla's post or Plato's.

    -Dan
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  7. #7
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    Quote Originally Posted by Plato View Post
    That is not how it works here.
    You should understand that this is not a tutorial service.
    Perhaps you need to hire a tutor.

    Or else you could post some of your own work on this problem or explain what you do not understand about the question.

    You are responsible for your own learning.
    Ok, I understand.

    I made up a simpler example that I tried to solve:

    f(x^2)=x
    f(f(x))=?

    So, looking at other examples,
    I got to f(x)=\sqrt{x}
    And now I have f(\sqrt{x})=?
    This is the stage where I completely freeze... I can't continue.
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  8. #8
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    Quote Originally Posted by ikislav View Post
    I made up a simpler example that I tried to solve:
    f(x^2)=x
    f(f(x))=?
    So, looking at other examples,
    I got to f(x)=\sqrt{x}
    And now I have f(\sqrt{x})=?
    This is the stage where I completely freeze... I can't continue.
    Actually your Ďmade upí has several difficulties with it.
    The original is much easier to work with because 2x-1 is a bijection.
    But letís modify your example, f(x^3)=x.
    Then  f(x) = \sqrt[3]{x} will work.
    Now f(f(x))=\sqrt[3]{{\sqrt[3]{x}}} = \sqrt[9]{x}.
    That works because \sqrt[3]{{\sqrt[3]{x}}} = \left( {\left( x \right)^{\frac{1}{3}} } \right)^{\frac{1}{3}}
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  9. #9
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    Quote Originally Posted by Plato View Post
    Actually your Ďmade upí has several difficulties with it.
    The original is much easier to work with because 2x-1 is a bijection.
    But letís modify your example, f(x^3)=x.
    Then  f(x) = \sqrt[3]{x} will work.
    Now f(f(x))=\sqrt[3]{{\sqrt[3]{x}}} = \sqrt[9]{x}.
    That works because \sqrt[3]{{\sqrt[3]{x}}} = \left( {\left( x \right)^{\frac{1}{3}} } \right)^{\frac{1}{3}}
    Ok... so:

    f(2x-1)=x

    t=2x-1

    x=\frac{t+1}{2}

    f(x)=\frac{x+1}{2}

    f(f(x))=f(\frac{x+1}{2})

    f(\frac{x+1}{2}) = \frac{\frac{x+1}{2}}{2}+\frac{1}{2}=

    =\frac{x+1}{2}*\frac{1}{2}+\frac{1}{2}=

    =\frac{x+1}{4}+\frac{2}{4}=

    =\frac{x+1+2}{4}=\frac{x+3}{4}

    f(f(x))=\frac{x+3}{4}

    Is this right?
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  10. #10
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    Quote Originally Posted by ikislav View Post
    Ok... so:
    f(f(x))=f(\frac{x+1}{2})
    Yes it is correct.
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  11. #11
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    Quote Originally Posted by Plato View Post
    Yes it is correct.
    Woohoo!

    Thanks for not making this easy, I feel large ammount of sattisfaction now.
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