If f(x) = a+bx, what are the real values of a and b such that f(f(f(1))) = 14 and f(f(f(0))) = -13.
+23246 If f(x) = a + bx --->
f(1) = a + b(1) = a + b
f( f(1) ) = f( a + b ) = a + b(a + b) = a + ab + b2
f( f( f(1) ) ) = f( a + ab + b2 ) = a + b( a + ab + b2 ) = a + ab + ab2 + b3 = 14
f(0) = a + b(0) = a
f( f(0) ) = a + b(a) = a + ab
f( f( f(0) ) ) = a + b(a + ab) = a + ab + ab2 = -13
Substituting a + ab + b2 = -13 into a + ab + ab2 + b3 = 14 ---> -13 + b3 = 14
---> b3 = 27 ---> b = 3
Since a + ab + ab2 + b3 = 14 ---> a + 3a + 9a + 27 = 14 ---> 13a + 27 = 14
---> a = -1
