Question 1187394

{{{f(x)=(x+1)^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+1)^2}}}
{{{x+1=0}}}
{{{x=-1}}}

a domain on which f is one-to-one and non-decreasing is all {{{x>=-1}}}

[{{{-1}}},{{{infinity}}})


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

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

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

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

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

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

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


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