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Math Help - composition functions

  1. #1
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    composition functions

    1. Find the Inverse of:

    f(x)=\frac{2x-1}{x-6}

    My answer was:


    f(x)=\frac{6x-1}{x-2}

    Is this Correct?


    2. Find f{\circ}g and g{\circ}f when
    f(x)=x^3-1 and g(x)=\sqrt[3]{x=1}

    I got:

    (x+1)^{1/18}-1


    \text{Thanks for the Help!!!}
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  2. #2
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    Quote Originally Posted by qbkr21 View Post
    1. Find the Inverse of:

    f(x)=\frac{2x-1}{x-6}

    My answer was:


    f(x)=\frac{6x-1}{x-2}

    Is this Correct?
    Yes, but this would look better if you did not use f for both functions, so you would have:

    f(x)=\frac{2x-1}{x-6}

    and its inverse as:

    g(x)=\frac{6x-1}{x-2}

    Also as a personel preference I would use a different variable name in this second definition so I would write:

    g(y)=\frac{6y-1}{y-2}

    is the inverse function of

    f(x)=\frac{2x-1}{x-6}.

    RonL
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  3. #3
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    Quote Originally Posted by qbkr21 View Post
    2. Find f{\circ}g and g{\circ}f when
    f(x)=x^3-1 and g(x)=\sqrt[3]{x+1}
    <br />
(f \circ g)(x)=(g(x))^3-1=(\sqrt[3]{x+1})^3-1=(x+1)-1=x<br />

    and:

    <br />
(g \circ f)(x)=\sqrt[3]{f(x)+1}=\sqrt[3]{x^3-1+1}=x<br />

    RonL
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    Re:

    Ok, but I am definatley did get those as answers I got new answers but noone of yours

    1. f{\circ}g= \sqrt{x+1}

    2. g{\circ}f= \sqrt[6]{x^3}


    PURE CALCULATOR

    \text{In your spare time could you please tell me what I did wrong}
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  5. #5
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    Quote Originally Posted by qbkr21 View Post
    ...
    2. Find f{\circ}g and g{\circ}f when
    f(x)=x^3-1 and g(x)=\sqrt[3]{x=1}

    I got:

    (x+1)^{1/18}-1


    \text{Thanks for the Help!!!}
    Hello,

    first I assume that you used the shift key where you better shouldn't use it. That means your problem reads:
    f(x)=x^3-1 and g(x)=\sqrt[3]{x+1}

    f{\circ}g means you have to calculate: f(g(x)). Therefore you have to plug the term of the function g in the place of the x in function f:

    f{\circ}g=f(g(x))=(\sqrt[3]{x+1})^3-1=x+1-1=x. That's the result CaptainBlack has already told you.

    Same procedure:

    g{\circ}f=g(f(x))=\sqrt[3]{x^3-1+1}=\sqrt[3]{x^3}=x. That's the result CaptainBlack has already told you.

    To be honest: I can't guess what you have done.

    EB
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  6. #6
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    Quote Originally Posted by qbkr21 View Post
    Ok, but I am definatley did get those as answers I got new answers but noone of yours

    1. f{\circ}g= \sqrt{x+1}

    2. g{\circ}f= \sqrt[6]{x^3}


    PURE CALCULATOR

    \text{In your spare time could you please tell me what I did wrong}
    Earboth has explained in more detail how we think this goes, if you want
    us to tell you where you went wrong, you will need to describe what you
    did.

    RonL
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    Re:

    I must have slipped up on my Calculator, thanks for the advice...
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  8. #8
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    Quote Originally Posted by qbkr21 View Post
    I must have slipped up on my Calculator, thanks for the advice...
    How are you doing this on a calculator? And surely your professor wants you to work out how you got your answer?

    -Dan
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    RE: Same Answers

    I keeps getting the same answers, so here is my work:

    1.
    f(g(x))= {(\sqrt[3]{x+1})^3}-1
    = {(x+1)^{1/18}}-1

    \text{and}


    2.
    g(f(x))= \sqrt[3]{(x^3-1)+1}
    = (x^3)^1/6
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    Re:

    Texas Instruments Voyage 200 the bling bling of All Calc's
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  11. #11
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    Quote Originally Posted by qbkr21 View Post
    I keeps getting the same answers, so here is my work:

    1.
    f(g(x))= {(\sqrt[3]{x+1})^3}-1
    = {(x+1)^{1/18}}-1

    \text{and}
    The power rule is: (x^a)^b = x^{ab}

    So
    \left ( \sqrt[3]{x+1} \right )^3 = \left ( (x + 1)^{1/3} \right )^3 = (x + 1)^{\frac{1}{3} \cdot 3} = x + 1

    Quote Originally Posted by qbkr21 View Post
    2.
    g(f(x))= \sqrt[3]{(x^3-1)+1}
    = (x^3)^{1/6}
    For the same reason as above:
    \sqrt[3]{(x^3-1)+1} = \sqrt[3]{x^3} = (x^3)^{1/3} = x^{3 \cdot \frac{1}{3}} = x

    -Dan
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  12. #12
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    Quote Originally Posted by qbkr21 View Post
    I keeps getting the same answers, so here is my work:

    1.
    f(g(x))= {(\sqrt[3]{x+1})^3}-1
    = {(x+1)^{1/18}}-1

    ...
    Hello,

    from your result I believe that you used the sqrt( command. Then you have calculated:

    \left( \left( (x+1)^\frac{1}{2}\right)^\frac{1}{3}\right)^\frac{  1}{3} , which will indeed give your result.

    Type on your calculator:
    Code:
    ((x+1)^(1/3))^3-1

    You'll get the correct result.

    EB
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    Re:

    Ok great i got x for my answer, but if you were a teacher grading tests would you accept the answer given? Do you think he or she would?
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  14. #14
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    Quote Originally Posted by qbkr21 View Post
    Ok great i got x for my answer, but if you were a teacher grading tests would you accept the answer given? Do you think he or she would?
    If you are speaking of the answers here:

    Quote Originally Posted by qbkr21 View Post
    I keeps getting the same answers, so here is my work:

    1.
    f(g(x))= {(\sqrt[3]{x+1})^3}-1
    = {(x+1)^{1/18}}-1

    \text{and}


    2.
    g(f(x))= \sqrt[3]{(x^3-1)+1}
    = (x^3)^{1/6}
    No.

    {(x+1)^{1/18}}-1 = \sqrt[18]{x+1} - 1

    and

    (x^3)^{1/6} = \sqrt[6]{x^3} = \sqrt{x}

    which are not the same as the correct answers.

    -Dan
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    Re:

    fine
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