## Calculus: Early Transcendentals 8th Edition

$f^{-1}(3)=1$ $f(f^{-1}(2))=2$
Let $f^{-1}(3)=a, f(a)=3,$ $a^3+a^2+a=3$ try $a=1$, LHS=$1^3+1^2+1=3$=RHS $\therefore f^{-1}(3)=1$ Let $f^{-1}(2)=b, f(b)=2,$ $\therefore f(f^{-1}(2))=f(b)=2$