# ono-one function question and partial fractions question

• Oct 1st 2011, 08:34 PM
mikel03
ono-one function question and partial fractions question
Im trying to solve this question but im unsure which formula to use.

a and b.

Attachment 22461
• Oct 1st 2011, 09:06 PM
TheChaz
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...
• Oct 1st 2011, 09:14 PM
mikel03
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
• Oct 1st 2011, 10:40 PM
CaptainBlack
Quote:

Originally Posted by mikel03
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
• Oct 1st 2011, 11:51 PM
mikel03
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
• Oct 2nd 2011, 12:01 AM
Siron
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.
• Oct 2nd 2011, 12:31 AM
mikel03
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
• Oct 2nd 2011, 12:33 AM
Siron
Re: ono-one function question and partial fractions question
Quote:

Originally Posted by mikel03
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?
• Oct 2nd 2011, 12:59 AM
mikel03
Re: ono-one function question and partial fractions question
i have no idea can you guide me through it step by step im confused
• Oct 2nd 2011, 01:23 AM
Siron
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.
• Oct 2nd 2011, 02:24 AM
mikel03
Re: ono-one function question and partial fractions question
what do u mean simultaneous equations. man im so confused
• Oct 2nd 2011, 02:40 AM
CaptainBlack
Re: ono-one function question and partial fractions question
Quote:

Originally Posted by mikel03
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
• Oct 2nd 2011, 06:07 AM
HallsofIvy
Re: ono-one function question and partial fractions question
Quote:

Originally Posted by Siron
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= __

Quote:

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

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.