- #1
synthetic.
- 12
- 0
Hello people. This is a problem from an A-Level Math book (though i am independantly learning, have given myself 2 months to cram the whole book down my throat :-S ), so, apparently, it is simple. I understand the principles behind functions, this is more of a problem with understanding complex fractions, i think. Well, here is the question as it is written:
A function f is defined on the set of real numbers by
f(x)= [tex]\frac{1-x}{x}[/tex] , (x [tex]\neq[/tex] 0)
Find, in its simplest form, an expression for f(f(x))
My Attempt:
f(f(x)) = f ( [tex]\frac{1-x}{x}[/tex] ) = ( 1-( [tex]\frac{1-x}{x}[/tex] )) / ( [tex]\frac{1-x}{x}[/tex] )
. . .as you can see, i am rubbish. I know that i am failing to see something. Maybe the following, which will give a further insight to my understanding (or mis), will help someone cure me.
[tex]\frac{1-x}{1}[/tex] = [tex]\frac{1}{x}[/tex](1-x) = [tex]\frac{1}{x}[/tex] -1
That is not correct, is it? Where am i going off the track in my understanding?
One last thing, the textbook i am studying has, allegedly, various errors in it, the answer it gives is [tex]\frac{2x-1}{1-x}[/tex] . I can only assume it is right, for now. I just do not see how that answer is arrived at, given the problem.
A function f is defined on the set of real numbers by
f(x)= [tex]\frac{1-x}{x}[/tex] , (x [tex]\neq[/tex] 0)
Find, in its simplest form, an expression for f(f(x))
My Attempt:
f(f(x)) = f ( [tex]\frac{1-x}{x}[/tex] ) = ( 1-( [tex]\frac{1-x}{x}[/tex] )) / ( [tex]\frac{1-x}{x}[/tex] )
. . .as you can see, i am rubbish. I know that i am failing to see something. Maybe the following, which will give a further insight to my understanding (or mis), will help someone cure me.
[tex]\frac{1-x}{1}[/tex] = [tex]\frac{1}{x}[/tex](1-x) = [tex]\frac{1}{x}[/tex] -1
That is not correct, is it? Where am i going off the track in my understanding?
One last thing, the textbook i am studying has, allegedly, various errors in it, the answer it gives is [tex]\frac{2x-1}{1-x}[/tex] . I can only assume it is right, for now. I just do not see how that answer is arrived at, given the problem.