The Real And Complex Number Systems

(i) If x ∈ R, it is clear that (x, x) ∈ S. So, S is reflexive.
(ii) If (x, y) ∈ S, it is clear that (y, x) ∈ S. So, S is symmetric.
(iii) If (x, y) ∈ S and (y, z) ∈ S, it is clear that (x, z) ∈ S. So, S is
transitive.
2.3 The following functions F and G are defined for all real x by the
equations given. In each case where the composite function G ◦ F can be
formed, give the domain of G◦F and a formula (or formulas) for (G ◦ F ) (x) .
(a) F (x) = 1 − x, G (x) = x 2 + 2x
Proof: Write
G ◦ F (x) = G [F (x)] = G [1 − x] = (1 − x) 2 + 2 (1 − x) = x 2 − 4x + 3.
It is clear that the domain of G ◦ F (x) is R.
(b) F (x) = x + 5, G (x) = |x| /x if x ≠ 0, G (0) = 0.
Proof: Write
G ◦ F (x) = G [F (x)] =
{
G (x + 5) =
|x+5|
if x ≠ −5.
x+5
0 if x = −5.
It is clear that the domain of G ◦ F (x) is R.
{ { 2x, if 0 ≤ x ≤ 1
x
(c) F (x) =
G (x) =
2 , if 0 ≤ x ≤ 1
1, otherwise,
0, otherwise.
Proof: Write
⎧
⎨ 4x 2 if x ∈ [0, 1/2]
G ◦ F (x) = G [F (x)] = 0 if x ∈ (1/2, 1]
⎩
1 if x ∈ R − [0, 1]
It is clear that the domain of G ◦ F (x) is R.
Find F (x) if G (x) and G [F (x)] are given as follows:
(d) G (x) = x 3 , G [F (x)] = x 3 − 3x 2 + 3x − 1.
Proof: With help of (x − 1) 3 = x 3 − 3x 2 + 3x − 1, it is easy to know
that F (x) = 1 − x. In addition, there is not other function H (x) such that
G [H (x)] = x 3 − 3x 2 + 3x − 1 since G (x) = x 3 is 1-1.
(e) G (x) = 3 + x + x 2 , G [F (x)] = x 2 − 3x + 5.
3
.