Question 1003177
{{{f(x)=sqrt(x^2+3x-4)/sqrt( (x-2)^2 ) }}}
The restrictions to the domain are
denominators cannot be zero, so {{{sqrt( (x-2)^2 )<>0}}}<-->{{{x<>2}}} , and
expressions inside square roots cannot be negative, so {{{x^2+3x-4>=0}}} .
(We know that {{{(x-2)^2}}} is not negative).
{{{x^2+3x-4=(x-1)(x+4)}}} is negative for {{{-4<x<1}}} ,
so ((-4,1) is not part of the domain,
and we knew that {{{x=2}}} is not part of the domain, either.
For any other value of {{{x}}} , {{{f(x)}}} is defined, so
the domain of  {{{f(x)}}} is {{{"("}}}{{{-infinity}}}{{{", -4 ) U [ 1 , 2 ) U ( 2 ,"}}}{{{infinity}}}{{{" )"}}} .
 
The function cannot be negative, because square roots are non-negative in their domain.
We know {{{f(x)=0}}} for {{{system(x=-4,"and",x=1)}}} .
{{{lim(x->2 , sqrt(x^2+3x-4)/sqrt((x-2)^2) )}}}={{{lim(x->2,sqrt(2^2+3*2-4)/sqrt((x-2)^2) )}}}={{{lim(x->2,sqrt(4+6-4)/sqrt((x-2)^2) )}}}={{{lim(x->2,sqrt(6)/sqrt((x-2)^2) )}}}{{{"="}}}{{{"+"}}}{{{infinity}}}
Since {{{f(x)}}} is continuous in {{{"[ 1 , 2 )"}}} ,
{{{f(1)=0}}} , and {{{lim(x->2 , f(x) )}}}{{{"="}}}{{{"+"}}}{{{infinity}}} ,
{{{f(x)}}} takes all values from {{{0}}} to {{{"+"}}}{{{infinity}}} in  {{{"[ 1 , 2 )"}}} .
The range of {{{f(x)}}} in the interval {{{"[ 1 , 2 )"}}} is {{{"[ 0 ,"}}}{{{"+"}}}{{{infinity}}}{{{" )"}}} .
Whatever it does in the rest of its domain, it cannot be negative,
so the range of {{{f(x)}}} across all of its domain is {{{"[ 0 ,"}}}{{{"+"}}}{{{infinity}}}{{{" )"}}} .