حل تشریحی سوال شماره 5 ریاضی 3

(0/75)$y=\sqrt{x+4}-1\Rightarrow y+1\sqrt{x+4}\Rightarrow {\left(y+1\right)}^2=x+4\Rightarrow {\left(y+1\right)}^2-4=x\Rightarrow f^{-1}\left(x\right)={\left(x+1\right)}^2-4$

(0/5) $D_{f^{-1}}=R_f=\left[-1,+\infty \right)$