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Math Help - Function math problem.

  1. #1
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    Function math problem.

    Hello, again!

    Is given: f\left(\frac{2x-1}{x+1}\right)=x^2-3x+4\;\;\;\text{find:}\;\;f(1)

    ----------------------------------------------------------------------------
    I've done tasks like this one before, like: f(x-2)=x^3-2x-1\;\;\;,\;\;\text{find:}\;f(1)

    A)\;11
    B)\;9
    C)\;20
    D)\;31,5


    t=x-2\;\;=>\;x=t+2

    f(t)=(t+2)^3-2(t+2)-1

    f(t)=t^3+3t^2*2+3t2^2+2^3-2t-4-1

    f(t)=t^3+6t^2+10t+3

    f(x)=x^3+6x^2+10x+3

    f(1)=1^3+6*1^2+10*1+3

    f(1)=7+13

    f(1)=20


    The attempt at a solution:

    f\left(\frac{2x-1}{x+1}\right)=x^2-3x+4\;\;\;\text{find:}\;\;f(1)

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

    t(x+1)=2x-1

    tx+t=2x-1

    tx+t+1=2x

    As you can see, I'm stuck in the beginning...how to separate X in one side??
    Thank you!
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Re: Function math problem.

    By inspection, you need to substitute x=2 to find f(1).
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  3. #3
    MHF Contributor Siron's Avatar
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    Re: Function math problem.

    Quote Originally Posted by kreshnik View Post
    Hello, again!

    Is given: f\left(\frac{2x-1}{x+1}\right)=x^2-3x+4\;\;\;\text{find:}\;\;f(1)
    Solve: \frac{2x-1}{x+1}=1
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  4. #4
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    Re: Function math problem.

    Quote Originally Posted by Siron View Post
    Solve: \frac{2x-1}{x+1}=1
    I guess substituting t with 1 is easier.

    \frac{2x-1}{x+1}=1\;\;\rightarrow\;\;x=2

    f(1)=1^2-3*1+4

    f(1)=1+1

    f(1)=2


    A)\;0\;\;\;\;B)\;1\;\;\;\;C)\;2\;\;\;\;D)\;3

    Thank you very much everyone.
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