Question 1035668
(a) {{{f^-1(1) = 1^5+2*1^3+3*1+1 = 7}}}.
To find f(1), substitute f(1) into {{{f^(-1)(x)}}}

==> {{{1 = f^-1(f(1))=(f(1))^5+2*(f(1))^3+3*f(1)+1}}}
==> {{{0 = (f(1))^5+2*(f(1))^3+3*f(1) = (f(1))( (f(1))^4+2*(f(1))^2+3               )}}} ==> {{{highlight(f(1) = 0)}}}, since {{{(f(1))^4+2*(f(1))^2+3 > 0}}}.

(b)  {{{f(x[0]) = 1}}} ==> {{{x[0] = f^(-1)(1) = 7}}}.

(c)   {{{f^-1(y[0])=1}}} ==> {{{f^-1(y[0])=(y[0])^5+2*(y[0])^3+3*y[0]x+1 = 1}}}

==> {{{(y[0])^5+2*(y[0])^3+3*y[0] = 0}}} ==> {{{(y[0])( (y[0])^4+2*(y[0])^2+3               ) = 0}}}  ==> {{{highlight(y[0] = 0)}}}.  (This is basically the same computation we had in the latter half of part (a).