How Do You Find g(x) When f(g(x)) Equals a Different Function?

AI Thread Summary
To find g(x) when f(g(x)) equals a different function, start by setting f(g(x)) equal to the given function. The equation becomes (3y-1)/(2y+5) = (x+9)/(12x-11) after substituting g(x) with y. Solving this equation directly can be complex, but the goal is to express x in terms of y using the inverse function. The process involves tedious algebraic manipulation to isolate g(x). Ultimately, the key is to replace x in f(x) with g(x) and solve for g(x) to find the desired relationship.
Gshandle
Messages
2
Reaction score
0

Homework Statement



Find g(x) if f(x) = (3x-1)/(2x+5) and f(g(x)) = (x+9)/(12x-11)

Homework Equations


N/A, as far as I know

The Attempt at a Solution



I tried doing it as though g(x) = y and it turned out like this:
(3y-1)/(2y+5)=(x+9)/(12x-11)
I very quickly saw that that wouldn't work though, so I'm kind of lost. This is independent work in order for me to be able to go into pre-calc, so while I can ask a teacher it is difficult to find time, hence why I'm relying on you guys :) Help would be much appreciated!
 
Physics news on Phys.org
Think about it like this: you have f(x), and you have f(x) in the specific case when x=g(x), what you are looking for is what does x have to be in f(x) to give f(g(x)).

So, what does x have to be exchanged with in (3x-1)/(2x+5) to give (x+9)/(12x-11)
 
Well, I already got that part of the logic which is why I set g(x) equal to y, but it didn't work, so I don't understand how to get it to work.
 
Maybe you can find f^{-1}(y)??

That is, set

y=\frac{3x-1}{2x+5}

and try to find x in function of y.
 
Ok, I just went through the math and its a bit of a long haul. Should be fairly easy but its tedious.

What you want to do is take your f(x) function, replace x with g(x), set it equal to your f(g(x)) function and then solve for g(x).
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top