Question 968784

by definition, injective (or one-to-one)  means that every member of "A" has its own {{{unique}}} matching member in "B" 

To show that {{{f}}} is 1-1, you could show that

{{{f(x)=f(y)}}} => {{{x=y}}}

one example: suppose {{{f(x)=(x-3)/(x+2)}}}

then{{{(x-3)/(x+2)=(y-3)/(y+2)}}}

{{{(x-3)(y+2)=(y-3)(x+2)}}}

{{{ cross(xy)+2x-3y-cross(6)=cross( yx)+2y-3x-cross(6)}}}

{{{2x-3y=2y-3x}}}

{{{2x+3x=2y+3y}}}

{{{5x=5y}}}

{{{x=y}}}...so,{{{f(x)}}} is 1-1


in your case {{{f (x)=x^2-3}}} , so if {{{x^2-3=y^2-3}}} then  {{{f (x)}}} is 1-1

{{{x^2-3=y^2-3}}}

{{{x^2-3+3=y^2}}}

{{{x^2=y^2}}}

=>{{{x=y}}} or {{{-x=y}}}...so,{{{f(x)}}} is not 1-1, it is not injective (one-to-one) on its domain



In fact we can do a "Horizontal Line Test": 

{{{drawing( 600, 600, -10, 10, -10, 10,
circle(-2,1,.12),circle(2,1,.12),
 graph( 600, 600, -10, 10, -10, 10,1, x^2-3)) }}} 

as you can see, {{{x=-2}}} and {{{x=2}}} have same {{{y=1}}}