What is domain ?
The domain of a function is the set of all possible values of the independent variable.
In other words, the domain of f(x) is the set of all real values of x for which f(x) is defined or meaningful.
Square root of a number is considered to be imaginary, if the value inside the square root is negative.
Square root of a number is considered to be real, if the value inside the square root is positive or zero.
Consider the square root function √f(x).
To find the domain of √f(x), you have to find the values of x for which √f(x) is real. Then,
f(x) ≥ 0
Domain of √f(x) is the values of x such that f(x) ≥ 0.
Consider the square root function 1/√f(x).
To find the domain of 1/√f(x), you have to find the values of x for which 1/√f(x) is real and defined.
In this case, if f(x) = 0, then the denominator is zero and the given function is undefined.
So,
f(x) > 0
Domain of 1/√f(x) is the values of x such that f(x) > 0.
Find the domain of the following functions :
Example 1 :
f(x) = √(x - 2)
Solution :
x - 2 ≥ 0
x ≥ 2
The domain is all real values greater than or equal to 2.
That is, x ∈ [2, ∞).
Example 2 :
f(x) = 1/√(1 - x)
Solution :
Since √(1 - x) is in denominator, '1 - x' can not be equal to zero and also it must be greater than zero.
1 - x > 0
Subtract 1 from both sides.
-x > -1
Multiply both sides by -1 (In an inequality, if you multiply or divide both sides by the same negative value, you have to flip the inequality sign.
x < 1
The domain is all real values less than 1.
That is, x ∈ (-∞, 1).
Example 3 :
f(x) = √(4 - x2)
Solution :
4 - x2 ≥ 0
22 - x2 ≥ 0
(2 + x)(2 - x) ≥ 0
Assume (2 + x)(2 - x) = 0 and solve for x.
x = -2 or x = 2
If these two values of x are marked on the real number line, we will have three intervals.
(-∞, -2], [-2, 2], [2, +∞)
Check each interval with the inequality (2 + x)(2 - x) ≥ 0 by taking a random value in the intervals.
(-∞, -2] does not satisfy the inequality (2 + x)(2 - x) ≥ 0
[-2, 2] satisfies the inequality (2 + x)(2 - x) ≥ 0
[2, +∞) does not satisfy the inequality (2 + x)(2 - x) ≥ 0
So, the domain is [-2, 2].
Example 4 :
f(x) = √(4 - x) + 1/√(x2 - 1)
Solution :
The given function has two parts :
Part 1 is √(4 - x)
Part 2 is 1/√(x2 - 1)
Part 1 :
4 - x ≥ 0
Subtract 4 from both sides.
-x ≥ -4
Multiply both sides by -1.
x ≤ 4
x ∈ (-∞, 4] ----(1)
Part 2 :
x2 - 1 > 0
x2 - 12 > 0
(x + 1)(x - 1) > 0
Assume (x + 1)(x - 1) = 0 and solve for x.
x = -1 or x = 1
If these two values of x are marked on the real number line, we will have three intervals.
(-∞, -1), (-1, 1), (1, +∞)
Check each interval with the inequality (x + 1)(x - 1) ≥ 0 by taking a random value in the intervals.
(-∞, -1] satisfies the inequality (x + 1)(x - 1) > 0
(-1, 1) does not satisfy the inequality (x + 1)(x - 1) > 0
(1, +∞) satisfies the inequality (x + 1)(x - 1) > 0
So, the domain is (-∞, 1)U(1, +∞).
x ∈ (-∞, 1)U(1, +∞) ----(2)
From (1) and (2), domain of the given function is
x ∈ (-∞, -1)U(1, 4]
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