Thread: ono-one function question and partial fractions question

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

a and b.

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...

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

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

Hello CaptainBlack

im still having trouble understanding the concept can you please create a example question so i can visualize the problem

2) What you have to do is the following:
You have to find in the expression:



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



Go further (expand the numerator), afterwards you have to make a (small) system to determine A,B and C.

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

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?

i have no idea can you guide me through it step by step im confused

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



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



And so:

, and (why?)

Solve this simultaneous equations.

what do u mean simultaneous equations. man im so confused

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:

,





CB

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,

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= __

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



And so:

, and (why?)

Solve this simultaneous equations.