Question 1210376
<pre>   

find the domain and range of {{{f(x)}}}{{{""=""}}}{{{(x^""^"")/(sqrt(1 - (sqrt(-x))^2))}}}

{{{drawing(400,400,-2,2,-2,2,green(line(-1,-3,-1,3)),

graph(400,400,-2,2,-2,2,(x)/(sqrt(1 - (sqrt(-x))^2))))}}}

x cannot be positive but can be 0. The denominator is always positive.
So the value of f is never positive but can be 0 when x=0.

What is under the radical in the denominator must be positive, not negative or
0.  Remember that -x can never be negative.

{{{1-(sqrt(-x))^2}}}{{{"">""}}}{{{0}}}
{{{-(sqrt(-x))^2}}}{{{"">""}}}{{{-1}}}
{{{(sqrt(-x))^2}}}{{{""<""}}}{{{1}}}
{{{-x}}}{{{""<""{{{""<""}}}{{{1}}}
{{{x}}}{{{"">""}}}{{{-1}}}

Remember again that -x is non-negative. 

{{{-x}}}{{{""=""}}}{{{1}}}

{{{x}}}{{{""=""}}}{{{-1}}} and since x is never positive but can be 0,

the domain is {{{-1<x<=0}}}

We set the denominator = 0 to find the vertical asymptote:

{{{sqrt(1 - (sqrt(-x))^2)}}}{{{""=""}}}{{{0}}}

{{{1 - (sqrt(-x))^2}}}{{{""=""}}}{{{0}}}

{{{-(sqrt(-x))^2}}}{{{""=""}}}{{{-1}}}

{{{(sqrt(-x))^2}}}{{{""=""}}}{{{1}}}

Remember that -x is non-negative. 

{{{-x}}}{{{""=""}}}{{{1}}}

{{{x}}}{{{""=""}}}{{{-1}}}

So the vertical asymptote is {{{x}}}{{{""=""}}}{{{-1}}}, indicated by the green line.

The domain is {{{matrix(1,5,"(",-1,",",0,"]")}}}

The range is {{{matrix(1,5,"(",-infinity,",",0,"]")}}}

Edwin</pre>