FINDING DOMAIN OF RATIONAL FUNCTIONS

Domain is set of possible inputs. To find domain of a rational function, we have to understand that the denominator should never become zero.

The values which makes the denominator as zero, those values should be excluded from the domain.

State the domain as interval notation.

Problem 1 :

Solution :

To find the domain of the rational function given above, we equate the denominator to 0.

x - 5 = 0

x = 5

5 is value which will make the function as undefined, so this value should be excluded from the domain.

Domain is R - {5}

Required interval notation is (-∞, 5) U (5, ∞)

Problem 2 :

x-1x2-3x-108

Solution :

Factoring the denominator,

x2 - 3x - 108

x2 - 12x + 9x - 108

= x(x - 12) + 9(x - 12)

= (x + 9)(x - 12)

Equating the factors to 0, we get x = -9 and x = 12

The values -9 and 12 will make the function as 0. These values should be excluded from domain.

Domain as interval notation is (-∞, -9) U (-9, 12) U (12, ∞)

Problem 3 :

x+5x2-3x-28

Solution :

Factoring the denominator,

x2 - 3x - 28

x2 - 7x + 4x - 28

= x(x - 7) + 4(x - 7)

= (x + 4)(x - 7)

Equating the factors to 0, we get x = -4 and x = 7

The values -4 and 7 will make the function as 0. These values should be excluded from domain.

Domain as interval notation is (-∞, -4) U (-4, 7) U (7, ∞)

Problem 4 :

3x-19x2-6x+1

Solution :

= 9x2 - 6x + 1

= 9x2 - 3x - 3x + 1

= 3x (3x - 1) - 1(3x - 1)

= (3x - 1)(3x - 1)

Equating each factor to 0, we get

x = 1/3 and 1/3

1/3 will make the denominator 0. So, this alone should be excluded 

Domain as interval notation is (-∞, 1/3) U (1/3, ∞)

Problem 5 :

6x2-x-122x2-5x+3

Solution :

= 2x2 - 5x + 3

= 2x2 - 2x - 3x + 3

= 2x (x - 1) - 3(x - 1)

= (2x - 3)(x - 1)

Equating each factor to 0, we get

x = 3/2 and 1

3/2 and 1 will make the denominator 0. So, these two values  should be excluded 

Domain as interval notation is (-∞, 3/2) U (3/2, 1) U(1, ∞).

Problem 6 :

x2x2+5

Solution :

x2 + 5 = 0

x2 = -5

x = √-5 (not defined)

No values will make the denominator as 0.

So, domain is (-∞, ∞)

Problem 7 :

x+2x2+3x-10

Solution :

= x2 + 3x - 10

= x2 + 5x - 2x - 10

= x(x + 5) - 2(x + 5)

= (x - 2)(x + 5)

Equating each factor to 0, we get

x = 2 and x = -5

2 and -5 will make the denominator 0. So, these two values  should be excluded 

Domain as interval notation is (-∞, -5) U (-5, 2) U (2, ∞).

Problem 8 :

x+2x2-3x

Solution :

= x2 - 3x

= x(x - 3)

Equating each factor to 0, we get

x = 0 and x = 3

0 and 3 will make the denominator 0. So, these two values  should be excluded 

Domain as interval notation is (-∞, 0) U (0, 3) U (3, ∞).

Problem 9 :

5(2x-10)

Solution :

√(2x -10) > 0

2x - 10 > 0

2x > 10

x > 5

5 is the value should be excluded from the domain.

So, the domain is (5, ∞)

Problem 10 :

11x2-100

Solution :

√(x2 - 100) > 0

x2 - 100 > 0

x2  > 100

Take square root on both side

x > ±10

-10 and 10 should be excluded from the domain.

So, the domain is (-∞, -10) U (10, ∞)

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