Question 1084544

given {{{f(x)=sqrt(2x-1) }}}and {{{g(x)= x^2+1}}}, where {{{x >=0}}}

1.without finding the formula for {{{f^-1}}},find{{{ f^-1(5)}}}

Determining the value of {{{ f^-1(5)}}} is equivalent to determining the {{{x}}}
value when  {{{ f(x)=5}}}

{{{5=sqrt(2x-1) }}}

{{{5^2=(sqrt(2x-1) )^2}}}

{{{25=2x-1}}}

{{{25+1=2x}}}

{{{26/2=x}}}

{{{x=13}}}

{{{ f^-1(5)=13}}}


2.find {{{f^-1(x)}}} and {{{g^-1(x)}}}

{{{f(x)=sqrt(2x-1) }}}

{{{y=sqrt(2x-1) }}}.............swap {{{x}}} and {{{y}}}

{{{x=sqrt(2y-1) }}}.......solve for  {{{y}}}

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

{{{x^2=2y-1 }}}

{{{x^2+1=2y }}}

{{{(1/2)(x^2+1)=y }}}


{{{f^-1(x)=(1/2)(x^2+1)}}}


 {{{g(x)= x^2+1}}}

 {{{y= x^2+1}}}

 {{{x= y^2+1}}}

 {{{x-1= y^2}}}

{{{y=sqrt(x-1)}}}

{{{g^-1(x)=sqrt(x-1)}}}



3.give range of f^-1(x) and give the domain of g^-1(x)

{{{f^-1(x)=x^2+1}}}

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

({{{-infinity}}},{{{infinity}}})
range: { {{{y}}} element {{{R}}} : {{{y>=1}}} }
[{{{1}}},{{{infinity}}})



4.express {{{(g o f)(x)}}} in simplified form 

I usually convert {{{( f o g)(x)}}} to the more intuitive {{{ f (g(x))}}} :

 {{{ f (g(x))=f( x^2+1)}}}

{{{f( x^2+1)=sqrt(2( x^2+1)-1) }}}

{{{f( x^2+1)=sqrt(2x^2+2-1) }}}

{{{f( x^2+1)=sqrt(2x^2+1) }}}