Question 1017508
.
{{{f(x)=y}}}={{{x^2-12}}}
{{{f^(-1)}}}:  {{{x=y^2-12}}}
{{{y^2=x+12}}}
{{{y=sqrt(x+12)}}}
.
{{{For}}} {{{x=4}}}:
{{{y=sqrt(4+12)}}}={{{sqrt(16)=4}}}
.
{{{Since}}} {{{f^(-1)=4}}}, ({{{f o}}} {{{f^(-1)}}})={{{f(4)}}}
.
{{{f(4)=4^2-12=16-12=4}}}
.