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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: Sanguinne on March 26, 2013, 08:01:10 pm

Title: Functional Equation
Post by: Sanguinne on March 26, 2013, 08:01:10 pm: Another problem I need help with...

The function f satisfies the functional equation f(f(x))=x for the maximal domain of f. The rule for the function is

a) f(x) = x +1
b) f(x) = x - 1
c) f(x) = ( x - 1 )/( x + 1)
d) f(x) = ln(x)
e) f(x) = ( x + 1 )/( x - 1)
Title: Re: Functional Equation
Post by: b^3 on March 26, 2013, 08:25:12 pm: Try each option and see.
So we are looking for $f(f(x))$ , that is wherever you see $x$ , we are going to put back in the function, $f(x)$ .

A
$\begin{alignedat}{1}f\left(f\left(x\right)\right) & =\left(x+1\right)+1 \\ & =x+2 \\ & \neq x \end{alignedat}$

B
$\begin{alignedat}{1}f\left(f\left(x\right)\right) & =\left(x-1\right)-1 \\ & =x-2 \\ & \neq x \end{alignedat} $

C
$\begin{alignedat}{1}f\left(f\left(x\right)\right) & =\frac{\frac{x-1}{x+1}-1}{\frac{x-1}{x+1}+1} \\ & =\left(\frac{x-1-x-1}{x+1}\right)\div\left(\frac{x-1+x+1}{x-1}\right) \\ & =\frac{-2}{x+1}\div\frac{2x}{x-1} \\ & =-\frac{2}{x+1}\times\frac{x-1}{2x} \\ & =-\frac{1-x}{x\left(x+1\right)},\: x\neq-1 \\ & \neq x \end{alignedat} $

D
$\begin{alignedat}{1}f\left(f\left(x\right)\right) & =\ln\left(\ln\left(x\right)\right),\: x>0 \\ & \neq x \end{alignedat} $

E
$\begin{alignedat}{1}f\left(f\left(x\right)\right) & =\frac{\frac{x+1}{x-1}+1}{\frac{x+1}{x-1}-1} \\ & =\left(\frac{x+1+x-1}{x-1}\right)\div\left(\frac{x+1-x+1}{x-1}\right) \\ & =\frac{2x}{x-1}\div\frac{2}{x-1} \\ & =\frac{2x}{x-1}\times\frac{x-1}{2} \\ & =x,\: x\neq1 \end{alignedat} $
Which makes the answer E.
Title: Re: Functional Equation
Post by: Sanguinne on March 26, 2013, 09:05:51 pm: thanks

i guess i was just making careless mistakes when simplifying
Title: Re: Functional Equation
Post by: fred42 on March 27, 2013, 05:10:31 pm: It should be noted that if (f(x)) = x, then the graph must be an inverse of itself and should therefore be symmetrical about the line y = x. This can help minimise the number of options that need to be checked.