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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??

Verified Answer :

nobillionaireNobley

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

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

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

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