Question 1187951

{{{f(x)=x^2+1}}}, where {{{x>0}}} 

inverse

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

{{{y=x^2+1}}}.....swap variables

{{{x=y^2+1}}}........solve for {{{y}}}

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

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

=>{{{f^-1(x)=sqrt(x-1)}}}


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