Calculus 8th Edition

The solutions are $f^{-1}(3)=1.$ $f(f^{-1}(2))=2.$
By noticing that $f(1)=1^5+1^3+1=3$ we get by the definition of the inverse function that $f^{-1}(3)=1.$ For the second part we will use the following property: If $f(x)=y$ then $f^{-1}(y)=x$ so $f(f^{-1}(y))=f(x)=y$. Applying this to our problem: $$f(f^{-1}(2))=2.$$