Given f(x) = (3x+1)/(x-3),find and write in simplest form:

(a) f(f(b))
f(b)=(3b+1)/(b-3)
f(f(b))=
f[(3b+1)/(b-3)]
=[3(3b+1)/(b-3)+1]/[(3b+1)/(b-3)-3]
=[(9b+3+b-3)/(b-3)]/[(3b+1-3b+9)/(b-3)]
=[10b/(b-3)]/[10/(b-3)]
=b
(b) [f(x)-f(a)]/(x-a)
=[(3x+1)/(x-3)-(3a+1)/(a-3)]/(x-a)
=[(3x+1)(a-3)-(3a+1)(x-3)]/(x-a)(x-3)(a-3)
=[3ax-9x+a-3-3ax+9a-x+3]/(x-a)(x-3)(a-3)
=-10(x-a)/(x-a)(x-3)(a-3)
=-10/(x-3)(a-3)