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The question is:
f(x)=[tex] \sqrt{x+1} [/tex]
Show that (f·[tex] f^{-1} [/tex])(a)=a=([tex] f^{-1} [/tex]·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??

Respuesta :

nobillionaireNobley
[tex]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[/tex]

[tex](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[/tex]

Therefore, [tex](fof^{-1})(a) = (f^{-1}of)(a) = a[/tex]

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