Question

If $f:[-1,\infty )\to [1,\infty )$ is defined by $f\left(x\right)=|x+1|+1$, then the value of $f^{-1}\left(7\right)$ is

Answer 1

For $x\in [-1,\infty )$,
$f\left(x\right)=x+1+1=x+2$
Now,
$y=x+2\Rightarrow x=y-2\Rightarrow f^{-1}\left(x\right)=x-2\therefore f^{-1}\left(7\right)=5$