Question

The question is:
f(x)=$\sqrt{x+1}$
Show that (f·$f^{-1}$)(a)=a=($f^{-1}$·f)(a) for any a in the respective domains.

Can somebody explain the question to me? I'm not even sure what it means. What are respective domains??

Answer 1

$f(x) = \sqrt{x+1} \\ y = \sqrt{x+1} \ or \ y^2 = x+1 \ or \ x = y^2 - 1 \\ Therefore, \ f^{-1}(x)=x^2 - 1$

$(fof^{-1})(a) = f(f^{-1}(a))=f(a^2-1) \\ =\sqrt{a^2-1+1}=\sqrt{a^2}=a \\ \\ (f^{-1}of)(a)=f^{-1}(f(a))=f^{-1}(\sqrt{a+1})= \\ (\sqrt{a+1})^2-1 = a+1-1 = a$

Therefore, $(fof^{-1})(a) = (f^{-1}of)(a) = a$