Question 1179077
{{{f(x)=sqrt(x-1)}}}

table

{{{x}}}|{{{f(x)}}} 

{{{1}}}|{{{0}}} ->{{{f(x) = sqrt(1-1)=0}}}
{{{2}}}|{{{1}}} ->{{{f(x) = sqrt(2-1)=1}}}
{{{3}}}|{{{1.4}}} ->{{{f(x) = sqrt(3-1)=sqrt(2)=1.4}}}
{{{4}}}|{{{1.7}}} ->{{{f(x) = sqrt(4-1)=sqrt(3)=1.7}}}


inverse 

{{{0}}}|{{{1}}} 
{{{1}}}|{{{2}}} 
{{{1.4}}}|{{{3}}} 
{{{1.7}}}|{{{4}}} 


{{{drawing(600,600,-10,10,-10,10, 
circle(0,1,.12),locate(0,1,p(0,1)),
circle(1,2,.12),locate(1,2,p(1,2)),
circle(1.7,4,.12),locate(1.7,3,p(1.7,4)),

circle(1,0,.12),locate(1,0,p(1,0)),
circle(2,1,.12),locate(2,1,p(2,1)),
circle(3,1.4,.12),locate(3,1.4,p(3,1.4)),
circle(4,1.7,.12),locate(4,1.7,p(4,1.7)),
graph(600,600,-10,10,-10,10,11, sqrt(x-1)),

graph(600,600,-10,10,-10,10,(x^2 + 1)*sqrt(sin(7x))/sqrt(sin(7x) )),


red(locate(-4,3,f^-1(x)=x^2 + 1)), green(locate(6,2,f(x)=sqrt(x-1)))   )}}}



domain of {{{f(x)}}}:

{ {{{x}}} element {{{R }}}: {{{x>=1}}} }

interval notation:

[{{{1}}},{{{infinity}}})


range:
{ {{{f(x)}}} element{{{ R}}} : {{{f(x)>=0}}} } 

interval notation:

[{{{0}}},{{{infinity}}})


domain of {{{f^-1(x)}}}:

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

interval notation:

[{{{-infinity}}},{{{infinity}}})

range:

{ {{{f(x)}}} element{{{ R}}} : {{{f(x)>=1}}} } 

interval notation:

[{{{1}}},{{{infinity}}})