Question 1177051
 Find the domain and range of 

A): 

{{{y=1/(sqrt(3-2x-x^2))}}}

domain: exclude all values of x that make denominator equal to zero

{{{sqrt(3-2x-x^2)=0}}} ...will be if
{{{3-2x-x^2=0}}}
{{{3-3x+x-x^2=0}}}
{{{3(1-x)+x(1-x)=0}}}
{{{(3+x)(1-x)=0}}}

solutions:

{{{x=-3}}}
{{{x=1}}}

domain is: { {{{x}}} element {{{R}}} : {{{-3<x<1}}} }

range:

to  find the range of a rational function use the derivative
{{{(d/dx)(1/sqrt(3 - 2 x - x^2)) = (x + 1)/(-x^2 - 2 x + 3)^(3/2)}}} => if numerator equal to zero then {{{x=-1}}}

plug in your function and find {{{y}}}

{{{y=1/(sqrt(3-2(-1)-(-1)^2))=1/sqrt(3+2-1)=1/sqrt(4)=1/2}}}

so range is { {{{y}}} element {{{R}}} : {{{y >= 1/2}}} }



B): 

{{{y=1/(sqrt(x^2+2x+3))}}}-> as you can see, no matter what  value is {{{x}}}, denominator cannot be equal to zero

domain: {{{R}}} (all real numbers)

range:

{{{(d/dx)(1/(sqrt(x^2+2x+3)))=(-x-1)/(x^2+2x+3)^(3/2)}}} =>if numerator equal to zero then {{{x=-1}}}

{{{y=1/(sqrt((-1)^2+2*(-1)+3))=1/sqrt(1-2+3)=1/sqrt(2)}}}->


so, range is: { {{{y}}} element {{{R}}} : {{{ 0< y <=1/sqrt(2)}}} }