Question 909003
Question 908997
<pre>
The red graph below is the graph of {{{"f(x)" =  (1-x)/x}}}.
The green dotted line is the graph of the identity function {{{"I(x)" = x}}}

{{{graph(400,400,-4,4,-4,4,(1-x)/x,x*sqrt(sin(9x))/sqrt(sin(9x)))}}}

The inverse function f<sup>-1</sup> is the reflection of the graph
of f(x) in or across the identity line whose equation is I(x)=x, which
is the green dotted line above.

Let's draw the graph of  f<sup>-1</sup> = {{{1/(x+1)}}} (in blue) to see it it looks like
it really it is the reflections of the red graph f(x) in or across the identity
function, which is the green dotted line I(x)=x:

{{{graph(400,400,-4,4,-4,4,(1-x)/x,x*sqrt(sin(9x))/sqrt(sin(9x)),1/(x+1))}}}

Yes it does look like it is.  To show this algebraically we must find the
composition f&#8728;<sup>-1</sup> and also the composition  f<sup>-1</sup>&#8728;f(x) and show that in both cases 
we get the right side of the identity function I(x)=x, which is x.

f&#8728;f<sup>-1</sup>(x) = {{{(1-(1/(x+1)))/(1/(x+1))}}}

Multiply top and bottom by LCD of (x+1)

f&#8728;f<sup>-1</sup>(x) = {{{(1(x+1)-1)/1}}}

f&#8728;f<sup>-1</sup>(x) = {{{(x+1-1)/1}}}

f&#8728;f<sup>-1</sup>(x) = {{{x}}}

So it gives x, which is the right side of the identity function I(x)=x,
which is the green dotted line.

----------

But we also have to show that it refects into I(x)=x both ways.

f<sup>-1</sup>&#8728;f(x) = {{{1/((1-x)/x+1)}}}

Multiply top and bottom by LCD of x

f<sup>-1</sup>&#8728;f(x) = {{{x/(1-x+x)}}}

f<sup>-1</sup>&#8728;f(x) = {{{x/1}}}

f<sup>-1</sup>&#8728;f(x) = {{{x}}}

We have found that both f&#8728;f<sup>-1</sup>(x) and f<sup>-1</sup>&#8728;f(x) are equal to the right side 
of the identity function I(x) = x, and that proves that each is the reflection
of the other in the green dotted line which is the identity function I(x) = x,
because composition with each other gives the right side of I(x) = x, which is
just x.

Edwin</pre>