HOW TO FIND DOMIAN OF A RADICAL FUNCTION

For what values x are the following expressions defined  and find domain.

Example 1 :

√(x-2)

Solution :

y = √(x-2)

Let f(x)  =  (x-2)

Here, radicand x-2 is defined by non-negative number.

x – 2 ≥ 0

Add 2 on both sides, we get

x – 2 + 2 ≥ 0 + 2

x ≥ 2

So, the possible value of x is

x ≥ 2

The function is defined for x ≥ 2.

So, the required domain is [2, ∞).

Example 2 :

√(3–2x)

Solution :

y  =  √(3–2x)

f(x)  =  (3–2x)

Here, radicand  3–2x is defined by non-negative number.

So, domain f(x)  ≥  0

3–2x ≥ 0

Subtract 3 on both sides, we get

3–2x-3 ≥ 0–3

-2x ≥ -3

Dividing by -2 on both sides, change the inequality ≥ into ≤

x ≤ 3/2

So, the possible value of x is

x ≤ 3/2

The function is defined for x ≤ 3/2.

So, the required domain is (-∞, 3/2].

Example 3 :

1/(x+2)

Solution :

y  =  1/(x+2)

Let f(x)  =  √(x+2)

Here, radicand x + 2 in the denominator is defined for all positive values except 0.

To find domain, we can use f(x) > 0.

x+2 > 0

Subtract 2 on both sides, we get

x+2-2 > 0–2

x > -2

So, the possible value of x is

x > -2

The function is defined for x > -2

So, the required domain is (-2, ∞).

Example 4 :

1/√(5+2x)

Solution :

Let f(x)  =  1/√(5+2x)

Here, radicand 5+2x in the denominator is defined by non-negative number.

To find domain, we can use f(x) > 0.

5 + 2x > 0

Subtract 5 on both sides, we get

5 + 2x - 5 > 0 – 5

2x > -5

Dividing by 2 on both sides, we get

x > -5/2

The possible value of x is x > -5/2

So, the domain is (-5/2, ∞).

Example 5 :

√[x(x – 2)]

Solution :

Let f(x)  =  √[x(x – 2)]

Here, radicand  x(x – 2) is defined by non-negative numbers.

So set that expressions,

x(x – 2) ≥ 0

x  =  0 and x – 2  =  0

Now, we have to choose only positive.

So the possible values of x are

x ≤ 0 or x ≥ 2

The Domain is (-∞, 0] u [2, ∞)

Example 6 :

(1/√x) + (1/√(2 – x))

Solution :

y  =  (1/√x) + (1/√(2 – x))

Here, radicand x and 2 – x in the denominators is defined by the non-negative numbers.

So set that expressions,

x > 0

2 – x > 0

Subtract  2 on both sides, we get

2 – x – 2 > 0 – 2

- x > - 2

x < 2

Now, we have to choose only positive.

So, the possible values of x are

0 < x < 2

The domain is (0, 2).

Example 7 :

√(x² – 3x)

Solution :

y  =  √(x² – 3x)

Here, radicand  x² – 3x is defined by non-negative numbers.

So set that expression,

x² – 3x ≥ 0

x(x – 3) ≥ 0

x  =  0 and x – 3  =  0

Now, we have to choose only positive.

So, the possible values of x are

x ≤ 0 or x ≥ 3

The domain is (-∞, 0] u [3, ∞)

Therefore, the required solution is x ≤ 0 or x ≥ 3

Example 8  :

√[(x+3)/(x-1)]

Solution  :

y  =  √[(x+3)/(x-1)]

Here, radicand (x+3) and (x–1) in the numerator and denominator is defined by non-negative values of x and the denominator will not become 0.

Now, we have to choose only positive.

So, the possible values of x are

x ≤ -3 or x > 1

The domain is (-∞, -3] u (1, ∞)

Therefore, the required solution is x ≤ -3 or x > 1

Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.