Question 1200539
<pre>
Even though the graph of f(x) = x^2 - 6x + 2 is this:

{{{drawing(300.1,300.1,-10,10,-10,10, graph(300.1,300.1,-10,10,-10,10, x^2-6x+2)))}}}

since {{{0<=x<=3}}}, we only use the part between and on the vertical lines 
at x=0 (the y-axis) and x=3, which is this:

{{{drawing(300.1,300.1,-10,10,-10,10, graph(300.1,300.1,-10,10,-10,10, (x^2-6x+2)*
(sqrt(x)/sqrt(x))*(sqrt(3-x)/sqrt(3-x))

), circle(3,-7,.1), 

circle(0,2,.1) 
                    ))}}}

To find the inverse of f(x) = x<sup>2</sup> - 6x + 2  for 0 <u><</u> x <u><</u> 3

1. We replace f(x) by y,

{{{y = x^2-6x+2}}}, {{{0<=x<=3}}}

2. We replace x by y and replace y by x:

{{{x = y^2-6y+2}}}, {{{0<=y<=3}}}

We solve for y:

{{{-y^2+6y-2+x=0}}}, {{{0<=y<=3}}}

Multiply through by -1 to make squared term positive:

{{{y^2-6y+2-x=0}}}, {{{0<=y<=3}}}

Enclose the term that represents "c" in parentheses:

{{{y^2-6y+(2-x)=0}}}, {{{0<=y<=3}}}

Use the quadratic formula:

{{{y = (-(-6) +- sqrt((-6)^2-4*1*(2-x) ))/(2*1) }}} 
{{{y = (6 +- sqrt(36-4(2-x) ))/2 }}}
{{{y = (6 +- sqrt(36-8+4x))/2 }}}
{{{y = (6 +- sqrt(28+4x) )/2 }}}
{{{y = (6 +- sqrt(4(7+x)))/2 }}}
{{{y = (6 +- 2sqrt(7+x))/2 }}}

{{{y = (2(3 +- sqrt(7+x)))/2 }}}
{{{y = (cross(2)(3 +- sqrt(7+x)))/cross(2) }}}
{{{y = 3 +- sqrt(7+x) }}}

Now we work out the domain {{{0<=y<=3}}}
Replace y by what y equals:
{{{0<=3 +- sqrt(7+x)<=3}}}
Add -3 to all three sides
{{{-3<="" +- sqrt(7+x)<=0}}}
Since it is between -3 and 0 we use the negative sign
for the square root. 
{{{-3<=-sqrt(7+x)<=0}}}
That tells us that for the inverse we use the negative
sign for the same square root in the inverse:
{{{y = 3 - sqrt(7+x) }}}
Going back to the inequality, we square all three sides. 
Squaring each side involves multiplying each side by 
itself, which is a negative quantity in the first two, 
and 0 is multiplied by 0, so we reverse the 
inequality signs:
{{{(-3)^2>=(sqrt(7+x))^2>=(0)^2}}}
{{{9>=7+x>=0}}}
Turn it around
{{{0<=7+x<=9}}}
Add -7 to all three sides:
{{{-7<=x<=2}}}
That's the domain of the inverse function. So the inverse is
{{{y = 3 - sqrt(7+x) }}}
although we write f<sup>-1</sup>(x) for y:

{{{f^(-1)}}}{{{(x)}}}{{{""=""}}}{{{3 - sqrt(7+x)}}}{{{for}}}{{{-7<=x<=2}}}  <---ANSWER

Here is the graph of the inverse on the same set of axes (in blue):

{{{drawing(300.1,300.1,-10,10,-10,10, 
graph(300.1,300.1,-10,10,-10,10, (x^2-6x+2)*
(sqrt(x)/sqrt(x))*(sqrt(3-x)/sqrt(3-x))),

circle(3,-7,.1), 

circle(0,2,.1),  


graph(300.1,300.1,-10,10,-10,10, 12,14,(3-sqrt(7+x))*(sqrt(7+x)/sqrt(7+x))*(sqrt(2-x)/sqrt(2-x))))}}}

And you see that the inverse is the reflection of the original funcetion
across the identity line, whose equation is y = x  (where x and y are
identically equal and the identity line is the line that bisects the 1st and 3rd
quadrants (in green, dashed since it's not part of either graph).

{{{drawing(300.1,300.1,-10,10,-10,10, 
circle(3,-7,.1), 

circle(0,2,.1),

circle(-7,2.85,.1), 

circle(2,0,.2),  
circle(0,2,.15), 

 

graph(300.1,300.1,-10,10,-10,10, (x^2-6x+2)*
(sqrt(x)/sqrt(x))*(sqrt(3-x)/sqrt(3-x))),
graph(300.1,300.1,-10,10,-10,10,15,x*sqrt(sin(9x))/sqrt(sin(9x))),
graph(300.1,300.1,-10,10,-10,10,13,14,(3-sqrt(7+x))*(sqrt(7+x)/sqrt(7+x))*(sqrt(2-x)/sqrt(2-x))))}}}


Edwin</pre>