Question 1163830
<pre>
The graph is this half of a parabola.

{{{drawing(400,400,-3,7,-3,7,circle(1,-1,.1),
graph(400,400,-3,7,-3,7,(x^2-2x)*sqrt(x-1.1)/sqrt(x-1.1)))}}}

The inverse is its projection in the identity line whose equation is
y = x, the dotted line below

{{{drawing(400,400,-3,7,-3,7,circle(1,-1,.1),
graph(400,400,-3,7,-3,7,x*sqrt(sin(9x))/sqrt(sin(9x))),
graph(400,400,-3,7,-3,7,(x^2-2x)*sqrt(x-1.1)/sqrt(x-1.1)))}}}

{{{matrix(1,3,
"f(x)" = x^2 - 2x,",",  x>1)}}}

Change f(x) to y

{{{matrix(1,3,
y = x^2 - 2x,",",  x>1)}}}

Interchange x and y.  You can interchange them now or later. 
I prefer to do it now, but your teacher might prefer to wait
to interchange them.  It really doesn't make any difference.

{{{matrix(1,3,
x = y^2 - 2y,",",  y>1)}}}

Solve for y by first swapping left and right sides:

{{{matrix(1,3,
y^2 - 2y=x,",",  y>1)}}}

Get 0 on the right side by subtracting x from both sides:

{{{matrix(1,3,
y^2 - 2y-x=0,",",  y>1)}}}

Use the quadratic formula:

{{{matrix(1,3,
y=(-(-2) +- sqrt((-2)^2-4(1)(-x)))/(2(1)),",",  y>1)}}}

{{{matrix(1,3,
y=(2 +- sqrt(4+4x))/2,",",  y>1)}}}

{{{matrix(1,3,
y=(2 +- sqrt(4(1+x)))/2,",",  y>1)}}}

{{{matrix(1,3,
y=(2 +- 2sqrt(1+x))/2,",",  y>1)}}}

{{{matrix(1,3,
y=(2(1 +- sqrt(1+x)))/2,",",  y>1)}}}

{{{matrix(1,3,
y=(cross(2)(1 +- sqrt(1+x)))/cross(2),",",  y>1)}}}

{{{matrix(1,3,
y=1 +- sqrt(1+x),",",  y>1)}}}

Since y > 1, we must choose the + sign and discard the negative sign.
Then we don't need to keep the y > 1 because the equation guarantees
that y will always be greater than 1.

{{{y=1 + sqrt(1+x)}}}

Next we change y to f<sup>-1</sup>(x):

{{{f^(-1)}}}{{{(x)}}}{{{""=""}}}{{{1+sqrt(1+x)}}}

The graph of that is the green curve below, which is the projection 
of the original (red) graph into the identity line, whose equation is y = x.

{{{drawing(400,400,-3,7,-3,7,circle(1,-1,.1),circle(-1,1.04,.1),
graph(400,400,-3,7,-3,7,x*sqrt(sin(9x))/sqrt(sin(9x))),
graph(400,400,-3,7,-3,7,20,1+sqrt(1+x)),
graph(400,400,-3,7,-3,7,(x^2-2x)*sqrt(x-1.1)/sqrt(x-1.1)))}}}

Edwin</pre>