DOMAIN OF A SQUARE ROOT FUNCTION WORKSHEET

Find the domain of the following functions :

1. f(x) = √(x - 2)

2. f(x) = 1/√(1 - x)

3. f(x) = √(4 - x²)

4. f(x) = √(4 - x) + 1/√(x² - 1)

Answers

1. Answer :

f(x) = √(x - 2)

x - 2 ≥ 0

x ≥ 2

The domain is all real values greater than or equal to 2.

That is, x ∈ [2, ∞).

2. Answer :

f(x) = 1/√(1 - x)

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).

3. Answer :

f(x) = √(4 - x²)

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].

4. Answer :

f(x) = √(4 - x) + 1/√(x² - 1)

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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