Question 1152172

{{{ G(t)=1/(t^2-16) }}}

{{{t^2-16 =0}}}

{{{t^2-4^2=0}}}

{{{(t-4)(t+4)=0}}}

if {{{(t-4)=0}}}=>{{{t=4}}}
if {{{(t+4)=0}}}=>{{{t=-4}}}


domain:

{ {{{t}}} element of {{{R}}}: {{{t<>-4}}} and {{{t<>-4 }}} }


range:

since {{{ G(t)=y}}}

 {{{ y=1/(t^2-16) }}}
{{{ y(t^2-16) =1}}}
{{{ yt^2-16y =1}}}......to both sides add -1 
{{{ yt^2-16y -1=0}}}-> For a quadratic equation of the form {{{ax^2+bx+c=0}}},  the discriminant is {{{b^2-4ac}}}

For {{{a=y}}},{{{b=0}}},{{{c=-16y-1}}}:discriminant is 

{{{0^2-4y(-16y-1)>=0}}}...expand

{{{-4y(-16y)-4(-1)>=0}}}

{{{64y^2+4>=0}}}

{{{16y^2+1>=0}}}


it will be for {{{y>0}}} or {{{y<=1/16}}}


{ {{{y}}} element of {{{R}}}: {{{y<=1/16}}} or {{{y>0}}} }