Question 956406
<pre>
{{{f(x)}}}{{{""=""}}}{{{3+sqrt(x-5)}}}.

f(x) and y are the same thing. So write it as:

{{{y}}}{{{""=""}}}{{{3+sqrt(x-5)}}}

What's under the square root, x-5, cannot be negative so it must be
greater than or equal to 0, so we have:

{{{x-5}}}{{{"">=""}}}{{{0}}}
{{{x}}}{{{"">=""}}}{{{5}}}}

So the domain is the set of all x-values greater than or equal to 5.

{x | x &#8805; 5} or in interval notation: [5,{{{infinity}}})

Since the square root {{{sqrt(x-5)}}} is never negative we can write

{{{sqrt(x-5)}}}{{{"">=""}}}{{{0}}}

We can add 3 to both sides to make the right side of f(x) above:

{{{3+sqrt(x-5)}}}{{{"">=""}}}{{{3}}}

And since  

{{{y}}}{{{""=""}}}{{{3+sqrt(x-5)}}}

{{{y}}}{{{"">=""}}}{{{3)}}}

So the range is the set of all y-values greater than or equal to 3.

{y | y &#8805; 3} or in interval notation: [3,{{{infinity}}}).  
 
-----------------------
f&#8728;g(x) = {{{f(g(x)^"")}}}, so we substitute the right side of
g(x) for x in the right side of f(x).

f&#8728;g(x){{{""=""}}}{{{f(g(x)^"")}}}{{{""=""}}}{{{""=""}}}{{{3+sqrt(tan(x)-5)}}}.

Edwin</pre>