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

a and b.

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- Oct 1st 2011, 08:34 PMmikel03ono-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 PMTheChazRe: 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... - Oct 1st 2011, 09:14 PMmikel03Re: 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 - Oct 1st 2011, 10:40 PMCaptainBlackRe: Help Me Please!!
- Oct 1st 2011, 11:51 PMmikel03Re: 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 AMSironRe: ono-one function question and partial fractions question
2) What you have to do is the following:

You have to find $\displaystyle A,B \ \mbox{and} \ C$ in the expression:

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

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

$\displaystyle \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 AMmikel03Re: 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 AMSironRe: ono-one function question and partial fractions question
- Oct 2nd 2011, 12:59 AMmikel03Re: 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 AMSironRe: ono-one function question and partial fractions question
If you expand the numerator of the RHS then you get:

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

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

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

And so:

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

Solve this simultaneous equations. - Oct 2nd 2011, 02:24 AMmikel03Re: ono-one function question and partial fractions question
what do u mean simultaneous equations. man im so confused

- Oct 2nd 2011, 02:40 AMCaptainBlackRe: ono-one function question and partial fractions question
- Oct 2nd 2011, 06:07 AMHallsofIvyRe: ono-one function question and partial fractions question
Or, at this point, recognizing that, since the denominators are the same, the numerators must be the same,

$\displaystyle 3x+ 11= A(x^2+1)+ (Bx+ C)(x- 3)$

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:

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

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

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

And so:

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

Solve this simultaneous equations.