HOW TO FIND THE DOMAIN AND RANGE OF SQUARE ROOT FUNCTION

Find the domain and range for the square root functions given below.

Problem 1 :

f(x) = √(x - 4) - 10

Solution:

Finding domain :

√(x - 4) ≥ 0

x ≥ 4

The domain will start from 4 and continue with positive values upto infinity.

So, domain is [4, ∞)

Finding range :

Using the concept of transformation, the graph will move down 10 units from origin and there is reflection.

Then range will start from -10 and continue upto +∞.

So, range is [-10, ∞)

Problem 2 :

Solution:

Finding domain :

√(x + 12) ≥ 0

x ≥ -12

The domain will start from -12 and continue with positive values upto infinity.

So, domain is [-12, ∞)

Finding range :

Using the concept of transformation, the graph will move up 3 units from origin and there is reflection.

Then range will start from 3 and continue upto +∞.

So, range is [3, ∞)

Problem 3 :

f(x) = √(x + 3) + 2

Solution:

Finding domain :

√(x + 3) ≥ 0

x ≥ -3

The domain will start from -3 and continue with positive values upto infinity.

So, domain is [-3, ∞)

Finding range :

Using the concept of transformation, the graph will move up 2 units from origin and there is reflection.

Then range will start from 2 and continue upto +∞.

So, range is [2, ∞)

Problem 4 :

f(x) = 2√(x - 5) - 6

Solution:

Finding domain :

√(x - 5) ≥ 0

x ≥ 5

The domain will start from 5 and continue with positive values upto infinity.

So, domain is [5, ∞)

Finding range :

Using the concept of transformation, the graph will move down 6 units from origin and there is reflection.

Then range will start from -6 and continue upto - ∞.

So, range is [-6, ∞)

Problem 5 :

f(x) = -√(5x)

Solution:

Finding domain :

-√(5x) ≥ 0

Taking square on both sides.

5x ≥ 0

Dividing by 5 on both sides.

x ≥ 0

Finding range :

Outputs are only negative values.

f(x) ≤ 0

Problem 6 :

f(x) = √(x + 1)

Solution:

Finding domain :

√(x + 1) ≥ 0

Taking square on both sides.

x + 1 ≥ 0

Subtracting 1 on both sides

x ≥ -1

Finding range :

Outputs are only positive values.

f(x) ≥ 0

Problem 7 :

f(x) = -√(2x) + 2

Solution:

Finding domain :

-√(2x) ≥ 0

Taking square on both sides.

2x ≥ 0

Dividing by 2 on both sides.

x ≥ 0

Finding range :

Using the concept of transformation, the graph will move up 2 units up from origin.

Then range will start from 2 and continue upto - ∞.

So, range is (-∞, 2]

Problem 8 :

f(x) = √(x + 2) + 5

Solution:

Finding domain :

√(x + 2) ≥ 0

x ≥ -2

The domain will start from -2 and continue with positive values upto infinity.

So, domain is [-2, ∞)

Finding range :

Using the concept of transformation, the graph will move up 5 units from origin and there is reflection.

Then range will start from 5 and continue upto +∞.

So, range is [5, ∞)

Problem 9 :

f(x) = √(x - 4) - 6

Solution :

Finding domain :

√(x - 4) ≥ 0

Taking square on both sides.

x - 4 ≥ 0

Adding 4 on both sides

x ≥ 4

The domain will start from 4 and continue with positive values upto infinity.

So, domain is [4, ∞)

Finding range :

Using the concept of transformation, the graph will move down 6 units down from origin.

Then range will start from -6 and continue upto + ∞.

So, range is [-6, ∞)

Problem 10 :

f(x) = -√(x - 6) + 5

Solution:

Finding domain :

√(x - 6) ≥ 0

x ≥ 6

The domain will start from 6 and continue with positive values upto infinity.

So, domain is [6, ∞)

Finding range :

Using the concept of transformation, the graph will move up 5 units from origin and there is reflection.

Then range will start from 5 and continue upto - ∞.

So, range is (-∞, 5]