FINDING THE DOMAIN OF A RATIONAL FUNCTION WITH SQUARE ROOT
Domain is set of possible inputs.
Case 1 :
When we have square root function at the numerator, the value that we receive after applying possible inputs should never become negative.
Example :
Solution :
Case 2 :
When we have square root function at the denominator, the value that we receive after applying possible inputs should never become negative as well 0.
Example :
Solution :
So, domain is (-∞,-2) and (2, ∞).
Case 3 :
When we have square root function in both numerator and denominator, the value that we receive after applying possible inputs should never become negatives and the denominator should not become 0.
Example :
Solution :
So, domain is [2, ∞).
Find the domain of the functions given below.
Problem 1 :
Solution:
Equating the denominator to 0, we get
√(x + 2) - 3 = 0
√(x + 2) = 3
x + 2 = 32
x + 2 = 9
x = 7
x ≠ 7
x + 2 ≥ 0
x ≥ -2
Domain:
{x| x ∈ R, x ≥ -2, x ≠ 7}
Problem 2 :
Solution:
√(9 - x2) = 0
9 - x2 > 0
x2 < 9
x < ± 3
Decomposing into intervals, (-∞, -3)(-3, 3) and (3, ∞).
|
x = -4 ∈ (-∞, -3) 9 - (-4)2 > 0 False |
x = 0 ∈ (-3, 3) 9 - (0)2 > 0 True |
x = 4 ∈ (3, ∞) 9 - 42 > 0 False |
Domain:
(-3, 3)
Problem 3 :
Solution:
√(1 - x) ≥ 0
1 - x ≥ 0
-x ≥ -1
x ≤ 1
√(x - 5) > 0
x > 5
Decomposing into intervals, (-∞, 1)(1, 5) and (5, ∞).
|
x = 0 ∈ (-∞, 1) √(1 - x) ≥ 0 and √(x - 5) > 0 √(1 - 0) ≥ 0 and √(0 - 5) > 0 False |
x = 2 ∈ (1, 5) √(1 - x) ≥ 0 and √(x - 5) > 0 √(1 - 2) ≥ 0 and √(2 - 5) > 0 False |
x = 2 ∈ (5, ∞)
√(1 - x) ≥ 0 and √(x - 5) > 0
√(1 - 2) ≥ 0 and √(2 - 5) > 0
False
Domain:
Ø
Problem 4 :
Solution:
x ≠ 0
√(9 - x2) ≥ 0
9 - x2 ≥ 0
x2 ≤ 9
x ≤ ± 3
Decomposing into intervals, (-∞, -3)(-3, 0), (0, 3) and (3, ∞).
|
x = -4 ∈ (-∞, -3) √(9 - x2) ≥ 0 and x ≠ 0 √(9 - 16) ≥ 0 and x ≠ 0 False |
x = -2 ∈ (-3, 0) √(9 - x2) ≥ 0 and x ≠ 0 √(9 - 4) ≥ 0 and x ≠ 0 True |
|
x = 1 ∈ (0, 3) √(9 - x2) ≥ 0 and x ≠ 0 √(9 - 1) ≥ 0 and x ≠ 0 True |
x = 4 ∈ (3, ∞). √(9 - x2) ≥ 0 and x ≠ 0 √(9 - 16) ≥ 0 and x ≠ 0 False |
Domain:
[-3, 3]
Problem 5 :
Solution:
Equating the denominator to 0, we get
√(x2 - 4) - 3 = 0
√(x2 - 4) = 3
x2 - 4 = 9
x2 = 13
x = √13
x ≠ √13
x2 - 4 ≥ 0
x2 ≥ 4
x ≥ ± 2
Domain:
{x| x ∈ R, x ≥ 2 but x ≠ √13}
Problem 6 :
Solution:
√(x - 8) = 0
x - 8 = 0
x = 8
x ≠ 8
x - 8 > 0
x > 8
Domain:
{x| x ∈ R, x > 8 but x ≠ 8}
Problem 7 :
Solution:
√(x + 3) > 0
x + 3 > 0
x > -3
Domain:
{x| x ∈ R, x > -3}
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