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

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

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

(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)]/...

全部带入化简即可。
第一题f(b)=(3b+1)/(b-3)
那么把f(b)带入f(x)中即可，也就是f(b)=x带入到原式中。
第二题同理，全部带入化简