FINDING DOMAIN AND RANGE OF SQUARE ROOT FUNCTIONS

Problem 1 :

f(x) = √(4x)

Solution :

Finding domain :

√(4x) ≥ 0

Taking square on both sides.

4x ≥ 0

Dividing by 4 on both sides.

x ≥ 0

Finding range :

Outputs are only positive values.

f(x) ≥ 0

Problem 2 :

f(x) = 3√(x - 1)

Solution :

Finding domain :

√(x - 1) ≥ 0

Taking square on both sides.

x - 1 ≥ 0

Adding 1 on both sides

x ≥ 1

Finding range :

Outputs are only positive values.

f(x) ≥ 0

Problem 3 :

f(x) = √(x + 8) - 2

Solution :

Finding domain :

√(x + 8) ≥ 0

Taking square on both sides.

x + 8 ≥ 0

Subtracting 8 on both sides

x ≥ -8

The domain will start from -8 and continue with positive values upto infinity.

So, domain is [-8, ∞)

Finding range :

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

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

So, range is [-2, ∞)

Problem 4 :

f(x) = √x - 2

Solution :

Finding domain :

√x ≥ 0

Taking square on both sides.

x ≥ 0

The domain will start from 0 and continue with positive values upto infinity.

So, domain is [0, ∞)

Finding range :

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

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

So, range is [-2, ∞)

Problem 5 :

f(x) = -√(3x - 5) + 5

Solution :

In front of the square root function, we have negative sign, it shows the reflection. The square root function is added with 5, so move the graph will move up 5 units.

Finding domain :

-√(3x - 5) ≥ 0

Taking square on both sides.

3x - 5 ≥ 0

Add 5,

3x ≥ 5

Divide by 3 on both sides,

x ≥ 5/3

The domain will start from 5/3 and continue with positive values upto infinity.

So, domain is [5/3, ∞)

Finding range :

Using the concept of transformation, the graph will move up 5 units up from origin.

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

So, range is (-∞, -5]

Problem 5 :

Solution :

Finding domain :

√(x - 1) ≥ 0

x ≥ 1

The domain will start from 1 and continue with positive values upto infinity.

So, domain is [1, ∞)

Finding range :

Using the concept of transformation, the graph will move down 4 units from origin and there is reflection.

Then range will start from -4 and continue upto - ∞.

So, range is (-∞, -4]

Problem 6 :

f(x) = -3√(x + 7)+9

Solution :

There is a reflection.

Finding domain :

√(x + 7) ≥ 0

x ≥ -7

The domain will start from -7 and continue with positive values upto infinity.

So, domain is [-7, ∞)

Finding range :

Using the concept of transformation, the graph will move up 9 units from origin and there is reflection.

Then range will start from 9 and continue upto - ∞.

So, range is (-∞, 9]

Problem 7 :

f(x) = √(x² - 4)

Solution :

Finding domain :

√(x² - 4) ≥ 0

Take square on both sides,

x² - 4 ≥ 0

Add 4 on both sides

x² ≥ 4

x  ≥ ±2

Decomposing into intervals and selecting random values from the interval, we get

From the number line above, shaded region is the solution. So, the domain is (-∞, -2] U [2, ∞)

Finding range :

Set of possible outputs are all positive values.

So, range is [0, ∞)

Problem 8 :

f(x) = √(9 - x²)

Solution :

Finding domain :

√(9 - x²) ≥ 0

Take square on both sides,

9 - x² ≥ 0

Subtract 9 on both sides

-x² ≥ -9

Dividing by - on both sides.

x  ≤ ±3

 (-∞, -3], [-3, 3] and [3, ∞)

Decomposing into intervals and selecting random values from the interval

So, the domain is [-3, 3].

Finding range :

Set of possible outputs are all positive values.

So, range is [0, 3]