Question 1187395

{{{f(x)=(x-5)^2}}} -> a domain will be all real numbers {{{R}}} 

Find a domain on which {{{f }}}is one-to-one and {{{non-decreasing}}} will be restricted

{{{0=(x-5)^2}}}
{{{x-5=0}}}
{{{x=5}}}

a domain on which f is one-to-one and non-decreasing is all {{{x>=5}}}

[{{{5}}},{{{infinity}}})



Find the inverse of f restricted to this domain f^-1(x)=

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

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

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

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

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



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