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Math Help - ono-one function question and partial fractions question

  1. #1
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    ono-one function question and partial fractions question

    Im trying to solve this question but im unsure which formula to use.

    a and b.

    ono-one function question and partial fractions question-q10.png
    Last edited by CaptainBlack; October 1st 2011 at 10:42 PM. Reason: ono-one function
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  2. #2
    Super Member TheChaz's Avatar
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    Re: Help Me Please!!

    Here's a lot more help than your question warrants...

    a) one-to-one means if f(a) = f(b), then a = b
    i.e. if (a - 2)^5 = (b - 2)^5, then a = b.
    This is true after taking (fifth) roots.

    b). Look up partial fractions. Or just have wolfram alpha do it...
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    Re: Help Me Please!!

    Thanks TheChaz,

    Im still having trouble, how can i input the queston into wolfram alpha.

    I dont understand question 2 where it says partial functions = a/x-3 + bx+c/x2+1
    Last edited by mikel03; October 1st 2011 at 10:38 PM.
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    Grand Panjandrum
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    Re: Help Me Please!!

    Quote Originally Posted by mikel03 View Post
    Thanks TheChaz,

    Im still having trouble, how can i input the queston into wolfram alpha.

    I dont understand question 2 where it says partial functions = a/x-3 + bx+c/x2+1
    It means find values for a,b and c such that the right hand side equals the lefthand side.

    CB
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    Re: ono-one function question and partial fractions question

    Hello CaptainBlack

    im still having trouble understanding the concept can you please create a example question so i can visualize the problem
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  6. #6
    MHF Contributor Siron's Avatar
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    Re: ono-one function question and partial fractions question

    2) What you have to do is the following:
    You have to find A,B \ \mbox{and} \ C in the expression:

    \frac{3x+11}{(x-3)(x^2+1)}=\frac{A}{x-3}+\frac{Bx+C}{x^2+1}

    So make a common denominator in the RHS, like this:

    \frac{A}{x-3}+\frac{Bx+C}{x^2+1}=\frac{A(x^2+1)+(Bx+C)\(x-3)}{(x-3)(x^2+1)}=...

    Go further (expand the numerator), afterwards you have to make a (small) system to determine A,B and C.
    Last edited by CaptainBlack; October 2nd 2011 at 02:45 AM. Reason: improve latout
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    Re: ono-one function question and partial fractions question

    Hey Siron thansk for the reply i got this answer. please let me know if im correct or wrong lol like im assuming i am

    a(x^2+1) + (x^2+1) (Bx + C)

    Thanks
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  8. #8
    MHF Contributor Siron's Avatar
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    Re: ono-one function question and partial fractions question

    Quote Originally Posted by mikel03 View Post
    Hey Siron thansk for the reply i got this answer. please let me know if im correct or wrong lol like im assuming i am

    a(x^2+1) + (x^2+1) (Bx + C)

    Thanks
    Is this an answer? You have to find A,B and C ...
    What do you get if you expand the numerator of the RHS?
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  9. #9
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    Re: ono-one function question and partial fractions question

    i have no idea can you guide me through it step by step im confused
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  10. #10
    MHF Contributor Siron's Avatar
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    Re: ono-one function question and partial fractions question

    If you expand the numerator of the RHS then you get:

    \frac{3x+11}{(x-3)(x^2+1)}=\frac{A(x^2+1)+(Bx+C)(x-3)}{(x-3)(x^2+1)}\\ \\  \\ \phantom{xxxxxxxxxxxxxxx}=\frac{Ax^2+A+Bx^2+Cx-3Bx-3C}{(x-3)(x^2+1)}\\ \\ \\ \phantom{xxxxxxxxxxxxxxx}=\frac{(A+B)x^2+(C-3B)x+(A-3C)}{(x-3)(x^2+1)}

    You notice the denominator's are equal so the numerator's have to be equal therefore:

    3x+11=(A+B)x^2+(C-3B)x+(A-3C)

    And so:

    C-3B=3, A-3C=11 and A+B=0 (why?)

    Solve this simultaneous equations.
    Last edited by CaptainBlack; October 2nd 2011 at 02:47 AM. Reason: improve layout of LaTeX
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  11. #11
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    Re: ono-one function question and partial fractions question

    what do u mean simultaneous equations. man im so confused
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    Grand Panjandrum
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    Re: ono-one function question and partial fractions question

    Quote Originally Posted by mikel03 View Post
    what do u mean simultaneous equations. man im so confused
    He means solve the three of equations on the last line of his post:

    C-3B=3,

    A-3C=11

    A+B=0

    CB
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  13. #13
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    Re: ono-one function question and partial fractions question

    Quote Originally Posted by Siron View Post
    If you expand the numerator of the RHS then you get:

    [tex]\frac{3x+11}{(x-3)(x^2+1)}=\frac{A(x^2+1)+(Bx+C)(x-3)}{(x-3)(x^2+1)}[tex]
    Or, at this point, recognizing that, since the denominators are the same, the numerators must be the same,
    3x+ 11= A(x^2+1)+ (Bx+ C)(x- 3)
    Taking x= 3, that gives 3(3)+ 11= 20= A(10) so that A= __.
    Taking x= 0, 3(0)+11= 11= A+ C(-3)= 2- 3C so that C= __.
    Taking x= 1, 3(1)+ 11= 14= A(2)+(B+ C)(-2)= 4+ B- 3 so B= __

    \\ \\  \\ \phantom{xxxxxxxxxxxxxxx}=\frac{Ax^2+A+Bx^2+Cx-3Bx-3C}{(x-3)(x^2+1)}\\ \\ \\ \phantom{xxxxxxxxxxxxxxx}=\frac{(A+B)x^2+(C-3B)x+(A-3C)}{(x-3)(x^2+1)}

    You notice the denominator's are equal so the numerator's have to be equal therefore:

    3x+11=(A+B)x^2+(C-3B)x+(A-3C)

    And so:

    C-3B=3, A-3C=11 and A+B=0 (why?)

    Solve this simultaneous equations.
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