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$\displaystyle f(x)=x^3$ generally, if $\displaystyle g(x)$is invertible, then $\displaystyle g^{-1}(g(f(x)))=g^{-1}(\sqrt{2-x^3})=f(x)$
Thanks, one more I don't get is:
Originally Posted by Coolman Thanks, one more I don't get is: use the same method, let$\displaystyle x=g^{-1}(y),f(g(x))=f(g(g^{-1}(y)))=f(y)=(g^{-1}(y))^2$
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