Ta có:
g(x) = x2 - 1 => g(x - 1)] = (x - 1)2 - 1
= x2 - 2x + 1 - 1 = x2 - 2x
f(x) = x - 1 => f[g(x - 1)] = f(x2 - 2x)
= (x2 - 2x) - 1 = x2 - 2x - 1
f(x) = x - 1 => f(x) - 1 = (x - 1) - 1
= x - 1 - 1 = x - 2
g(x) = x2 - 1 => g[f(x) - 1] = g(x - 2)
= (x - 2)2 - 1 = x2 - 4x + 4 - 1
= x2 - 4x + 3
Vậy f[g(x - 1)] = x2 - 2x - 1 và g[f(x) - 1] = x2 - 4x + 3