HOW TO FIND DOMAIN OF A FUNCTION

Find the domain of the following functions :

Example 1 :

f(x) = 1/(x + 2)

Solution :

The given function accepts all real values except -2. In the given function if we apply -2 instead of x, it will become undefined.

Hence the domain of f(x) is R - {-2}.

Example 2 :

f(x) = (x - 1)/(x - 2)

Solution :

In numerator, we don't have any restriction for x values. But in the denominator, if we apply x = 2 the function will become undefined.

Hence the domain of f(x) is R - {2}.

Example 3 :

f(x) = (2x - 3)/(x² - 3x + 2)

Solution :

Since we have quadratic equation in the denominator, let us find factors.

x² - 3x + 2 = (x - 1)(x - 2)

f(x) = (2x - 3)/[(x - 1)(x - 2)]

In numerator, we don't have any restriction for x values. But in the denominator, if we apply x = 2 or x = 1 the function will become undefined.

Hence the domain of f(x) is R - {1, 2}.

Example 4 :

f(x) = (x² + 3x + 5)/(x² - 5x + 4)

Solution :

We could not find linear factors of the quadratic equation which is in the numerator.

x² - 5x + 4 = (x - 1)(x - 4)

f(x) = (x² + 3x + 5)/(x - 1)(x - 4)

In numerator, we don't have any restriction for x values. But in the denominator, if we apply x = 1 or x = 4 the function will become undefined.

Hence the domain of f(x) is R - {1, 4}.

Example 5 :

f(x) = (2x + 1)/(x²  - 9)

Solution :

f(x) = (2x + 1)/(x²  - 3²)

By comparing the denominator x²  - 3² with the algebraic identity a²  - b², we get two factors (x + 3)(x - 3).

f(x) = (2x + 1)/(x + 3)(x - 3)

In numerator, we don't have any restriction for x values. But in the denominator, if we apply x = -3 or x = 3 the function will become undefined.

Hence the domain of f(x) is R - {-3, 3}.

Example 6 :

f(x) = √(2x - 3)

Solution :

f(x) = √(2x - 3)

The given function is a square root function. If there is a negative value inside the square root, it is considered as imaginary value. So, we can not have a negative value inside the square root.

2x - 3 ≥ 0

2x ≥ 3

x ≥ 3/2

x ≥ 1.5

Hence the domain of f(x) is [1.5, +∞).

Example 7 :

f(x) = 1/√(x + 3)

Solution :

f(x) = 1/√(x + 3)

The given function is a square root function and also we have square root in denominator.

x = -3 will make the denominator zero and the function f(x) will become undefined. And also, since we have (x + 3) inside the square root, it has to be a positive value.

Finally, (x + 3) can not be equal to zero and  also it has to be a positive value.

x + 3 > 0

x > -3

Hence the domain of f(x) is (-3, +∞).

Example 8 :

f(x) = √(3x - 18)/√(x - 5)

Solution :

f(x) = √(2x - 3)/√(x - 5)

We have √(2x - 3) in numerator. So, (2x - 3) can be equal to zero or it has to be a positive value.

3x - 18 ≥ 0

3x ≥ 18

x ≥ 6 ----(1)

We have √(x - 5) in denominator. So, (x - 5) can not be equal to zero or it has to be a positive value.

x - 5 > 0

x > 5 ----(2)

The common region of (1) and (2) is [6, +∞).

Hence the domain of f(x) is [6, +∞).

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