Question 981349
You mean  {{{g(x)=sqrt(2x-1)}}} ?


If y is the inverse expression, then  {{{sqrt(2y-1)=x}}}.


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


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


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


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


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



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Domain and Range are now switched between {{{g(x)}}} and {{{g^-1(x)}}}.
Domain for g(x) is {{{x>=1/2}}}; range for g(x) is only non-negative numbers, so {{{g(x)>=0}}} range.
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{{{g^-1(x)}}} Domain & Range:
domain is {{{x>=0}}}, and range  {{{y>=1/2}}}.