Question 169749
Find the domain and range of the inverse of the function.
<pre><font size = 4 color="indigo"><b>
{{{matrix(1,3,   f(x) = 6-x^2,";", x >= 0)}}}  

Let's do it graphically first, and then we'll do it
algebraically.

First we'll draw the graph of {{{y=6-x^2}}}

{{{drawing(400,400,-10,10,-10,10,
locate(3.5,-5,y=6-x^2),

graph(400,400,-10,10,-10,10,(6-x^2) ))}}}

That is a parabola with vertex (0,6).
However because of the {{{x>=0}}}
restriction, we will have to eliminate
the left half of the graph to have the
graph of f(x), which is this:

{{{drawing(400,400,-10,10,-10,10,
locate(3.5,-5,f(x)),
graph(400,400,-10,10,-10,10,(6-x^2)*sqrt(x)/sqrt(x)) )}}}

The domain of {{{f(x)}}} is {{{matrix(1,5, "[", 0, ",", infinity, ")")}}}   
The range of {{{f(x)}}} is {{{matrix(1,5, "(", -infinity, ",", 6, "]")}}}

Now let's draw what is called "the identity line",
which is the 45°-line through the origin whose 
equation is {{{y=x}}}.  I'll draw the identity line
in green dotted:

{{{drawing(400,400,-10,10,-10,10,
locate(3.5,-5,f(x)),
graph(400,400,-10,10,-10,10,(6-x^2)*sqrt(x)/sqrt(x),
x*sqrt(sin(4x))/sqrt(sin(4x)) ))}}}

Now let's draw the graph of {{{matrix(1,2,f^(-1),(x))}}},
the inverse of {{{f(x)}}} by reflecting it across the 
green dotted identity line like this:

{{{drawing(400,400,-10,10,-10,10,
graph(400,400,-10,10,-10,10,(6-x^2)*sqrt(x)/sqrt(x),
x*sqrt(sin(4x))/sqrt(sin(4x)) ),
graph(400,400,-10,10,-10,10,sqrt(6-x) ),
locate(-7,5,f^(-1)), locate(-5.5,4.6,(x)), locate(3.5,-5,f(x))
)}}}

The domain of {{{f^(-1)(x)}}} is {{{matrix(1,5, "(", -infinity, ",", 6, "]")}}}
The range of {{{f^(-1)(x)}}} is {{{matrix(1,5, "[", 0, ",", infinity, ")")}}}


Notice that the range of {{{f(x)}}} is the domain of {{{f^(-1)}}}{{{(x)}}},
and the domain of {{{f(x)}}} is the range of {{{f^(-1)}}}{{{(x)}}}.
That's doing it graphically. Now let's do it
algebraically:

{{{matrix(1,3,   f(x) = 6-x^2,";", x >= 0)}}}

Replace {{{f(x)}}} by {{{y}}}

{{{matrix(1,3,   y = 6-x^2,";", x >= 0)}}}

Interchange {{{x}}} and {{{y}}}:

{{{matrix(1,3,   x = 6-y^2,";", y >= 0)}}}

Solve for y:

{{{matrix(1,3,   y^2 = 6-x,";", y >= 0)}}}

We take the square roots of both sides, and 
since {{{y>=0}}}, we do not need "±".

{{{matrix(1,3,   y = sqrt(6-x),";", y >= 0)}}}

Now we replace {{{y}}} by {{{f^(-1)}}}{{{(x)}}}

{{{matrix(1,6,  f^(-1), (x), "=", sqrt(6-x),";", y >= 0)}}}

What is under the square root radical must not be negative,
so {{{6-x>=0}}} or {{{-x>=-6}}} or {{{x<=6}}}

So, because of {{{x<=6}}},

The domain of {{{f^(-1)(x)}}} is {{{matrix(1,5, "(", -infinity, ",", 6, "]")}}}

and because of the {{{y>=0}}},

The range of {{{f^(-1)(x)}}} is {{{matrix(1,5, "[", 0, ",", infinity, ")")}}}

Edwin</pre>