DOMAIN OF A SQUARE ROOT FUNCTION

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.

 4

∈ (-∞, 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, +∞).

∈ (-, 1)U(1, +∞) ----(2)

From (1) and (2), domain of the given function is

∈ (-∞, -1)U(1, 4]

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