Question 930211
{{{ f(x)=(x+2)^2  }}}

to {{{ find f^-1  }}}  first recall that {{{ f(x)=y  }}} 


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


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


{{{sqrt(x)=sqrt((y+2)^2)  }}}


{{{sqrt(x)=y+2  }}}


 {{{y=(-2 +- sqrt(x ))}}}


so, {{{ find f^-1 =(-2 +- sqrt(x )) }}} 


{{{ graph( 600, 600, -10, 10, -10, 10, (x+2)^2,- sqrt(x)-2, sqrt(x)-2) }}}