Question 121381
{{{f(x)=x^2}}}
There are no restrictions on domain. 
The domain is ({{{-infinity}}},{{{infinity}}}).
Since the value of x is squared, f(x) will always be equal or greater than 0.
The range is [{{{0}}},{{{infinity}}}).
The bracket, [, means includes 0 in this definition. 
{{{f(x)=1/((x^2-1)^(1/2))}}}
There are several restrictions on the domain. 
The function is not defined when the denominator equals zero. 
Therefore x cannot equal 1 or -1. 
The square root also requires a positive value. 
{{{x^2-1>0}}}
{{{x^2>1}}}
{{{x>1}}} and {{{x<-1}}}
 
Since the output of the square root function is also always positive, 
the range is also positive. 
As x grows large, either negative or positive, f(x) goes to zero.
 The range is ({{{0}}},{{{infinity}}}).