Question on Function of Function.

Question

$$f(x)=\frac{x+2}{1-2x}$$ $$g(x)=\frac{2x+1}{2-x}$$

Find $$(fofofo...ofOgogo...og)=\frac{1}{x}$$ {fofo... are 101 times and gogo.. are 100 times}

Then Find $x$?

I calculated as follows Since $$(fog)=\frac{\frac{2x+1}{2-x}+2}{1-2{\frac{2x+1}{2-x}}}$$ $$(fog)=\frac{2x+1+4-2x}{2-x-4x-2}$$ $$(fog)=\frac{5}{-5x}=-\frac{1}{x}$$

and $$(fog)o(fog)=x$$

So 100th term as $(fog)o(fog)o...o(fog) = x$

Hence $$(fofofo...ofOfofo...og)=\frac{1}{x}$$ $$f(fog)o(fog)o(fog)o....o(fog)=1/x$$ $$f(x)=1/x$$ $$f(x)=\frac{x+2}{1-2x}=\frac{1}{x}$$ $$x^2+2x=1-2x$$ $${x^2}+4x-1=0$$

$$x=\frac{-4\pm\sqrt{4^2+4}}{2(1)}$$ $$x=-2\pm\sqrt{5}$$

Is the method and Answer Correct? Need Your Suggestion.

Answer 1

Your method is almost correct. The result is definitely correct.

You only need to show that $f$ and $g$ commute (which is true):

• $(f\circ g)(x) = (g\circ f)(x) = -\frac{1}{x}$

Hence, you have

• $((f\circ g)\circ(f\circ g))(x) = x$

Since $f$ and $g$ commute using the symbols $f^n = \underbrace{f\circ \cdots \circ f}_{n-fold\; composition\; of \;f}$ etc., you get your result

$f^{101}\circ g^{100} = ((f\circ g)^2)^{50}f = f$

The remaining part of your solution is exactly as you did.

Answer 2

Composition of functions is not -- in general -- a commutative process, so what you've computed is apparently different from what the problem specifies. You are not to rearrange the order of composition for different functions unless you've first shown that it is allowable.

What you need may be a little tenacity. Composing functions of the types above with themselves -- that is $ffff\cdots f(x)$ -- usually ends one up in the identity function after a sufficient number of iterated compositions. Thus, you need not compute after some point on.