2 Qs of Function

(1) f(x/2) = 3x/2 + 5
(a) f(1) = f(2/2) = 6/2 + 5 = 8
f(2) = f(4/2) = 12/2 + 5 = 11
(b) f(u) = f(2u/2) = 3(2u)/2 + 5 = 3u + 5
(2) f(2x) = x^2 + 3 - 4
(a) f(2) = f(2*1) = 1 + 3 - 4 = 0
f(4) = f(2*2) = 4 + 3 - 4 = 3
(b) f(x) = f[2(x/2)]
= (x/2)^2 + 3 - 4
= x^2/4 - 1
But I suspect the question should be:
(2) f(2x) = x^2 + 3x - 4
(a) f(2) = f(2*1) = 1 + 3 - 4 = 0
f(4) = f(2*2) = 4 + 6 - 4 = 6
(b) f(x) = f[2(x/2)]
= (x/2)^2 + 3(x/2) - 4
= x^2/4 + 3x/2 - 4