Solution: (3)
f (x) = [x / (1 + x2)]
1 + x2 ≠ 0
x2 ≠ -1
x ∈ R
Domain: x ∈ R
1 + x2 ≥ 1
f (x) = y = [x / (1 + x2)]
yx2 – x + y = 0
D ≥ 0
(-1)2 – 4yy ≥ 0
1 – 4y2 ≥ 0
4y2 – 1 ≤ 0
(2y – 1) (2y + 1) ≤ 0
y ∈ [-1 / 2, 1 / 2]
Solution: (3)
f (x) = [x / (1 + x2)]
1 + x2 ≠ 0
x2 ≠ -1
x ∈ R
Domain: x ∈ R
1 + x2 ≥ 1
f (x) = y = [x / (1 + x2)]
yx2 – x + y = 0
D ≥ 0
(-1)2 – 4yy ≥ 0
1 – 4y2 ≥ 0
4y2 – 1 ≤ 0
(2y – 1) (2y + 1) ≤ 0
y ∈ [-1 / 2, 1 / 2]