Question 1127951

the inverse of {{{f(x) = (x - 5)^2}}} for {{{x >= 5}}}

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

{{{y= (x - 5)^2}}} swap {{{x}}} and {{{y}}}

{{{x= (y - 5)^2}}}........solve for {{{y}}}

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

{{{sqrt(x)= y - 5}}}

{{{sqrt(x)+5= y }}}

so,function {{{g}}} is the inverse of function{{{ f}}} we have

 {{{g(x)=sqrt(x)+5}}} for {{{x >= 5}}}