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anonymous

  • one year ago

Find gof if f(x)=x+(1/x) and g(x)= (x+1)/(x+2)

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  1. princeharryyy
    • one year ago
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    g(f(x))= (f(x)+1)/(f(x)+2)

  2. princeharryyy
    • one year ago
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    just out the value of f(x) in place of f(x) in above equation n solve

  3. princeharryyy
    • one year ago
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    @Theanitrix

  4. princeharryyy
    • one year ago
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    put*

  5. anonymous
    • one year ago
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    I know that, I just am not sure how to solve it

  6. princeharryyy
    • one year ago
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    ok

  7. anonymous
    • one year ago
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    the fraction confuses me

  8. Jhannybean
    • one year ago
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    \[(g \circ f) = g\left(x+\frac{1}{x}\right) = \frac{\left(x+\dfrac{1}{x}\right)+1}{\left(x+\dfrac{1}{x}\right)+2}\]

  9. princeharryyy
    • one year ago
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    well, you got the answer now, I guess.

  10. Jhannybean
    • one year ago
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    That isn't the answer.

  11. princeharryyy
    • one year ago
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    that's almost the final answer @Jhannybean

  12. anonymous
    • one year ago
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    the answer would be (x^2+x+1)/(x+1)^2

  13. anonymous
    • one year ago
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    but i want to know how they got that answer

  14. Jhannybean
    • one year ago
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    multiply both the numerator and denominator by \(x\). \[\frac{\left(x+\dfrac{1}{x}\right)+1}{\left(x+\dfrac{1}{x}\right)+2} \cdot \frac{x}{x} \]

  15. princeharryyy
    • one year ago
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    (g∘f)=g(x+1/x)=(x^2+1+x)/(x^2+2+x)

  16. anonymous
    • one year ago
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    is that because it is the denominator of 1/x? @Jhannybean

  17. Jhannybean
    • one year ago
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    Yes @Theanitrix

  18. anonymous
    • one year ago
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    okay so would that same thing apply to fof?

  19. Jhannybean
    • one year ago
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    \[(f\circ f)= f\left(x+\frac{1}{x}\right) = \left(x+\frac{1}{x}\right)+\dfrac{1}{x+\dfrac{1}{x}}\]

  20. Jhannybean
    • one year ago
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    but when you're trying to solve this... set all that complicated junk = \(a\) or any other variable. find the common denominator, and simplify.

  21. Jhannybean
    • one year ago
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    \[\text{let a}=x+\frac{1}{x}\]\[(f\circ f) = a+\frac{1}{a} = \frac{a^2+1}{a} = \dfrac{\left(x+\dfrac{1}{x}\right)^2+1}{\left(x+\dfrac{1}{x}\right)} \]

  22. Jhannybean
    • one year ago
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    \[=\frac{x^2+\dfrac{1}{x^2}+2 +1}{\dfrac{x^2+1}{x}} =\frac{x^2+\dfrac{1}{x^2}+3}{\dfrac{x^2+1}{x}} \] So now you'd want to multiply both numerator and denominator by \(\dfrac{x}{x}\)

  23. anonymous
    • one year ago
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    i did it a different way

  24. anonymous
    • one year ago
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    and got the answer x^4+3x^2+1/x(x^2+1)

  25. Jhannybean
    • one year ago
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    Thats correct

  26. anonymous
    • one year ago
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    okay, i understood your approach as well so thank you

  27. Jhannybean
    • one year ago
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    \[\frac{x^2+\dfrac{1}{x^2}+3}{\dfrac{x^2+1}{x}} \cdot \frac{x}{x} =\frac{\dfrac{x^4+3x^2+1}{x}}{x^2+1} = \color{red}{\frac{x^4+3x^2+1}{x(x^2+1)}}\]

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