HOW TO FIND DOMAIN OF A RATIONAL FUNCTION

Find the domain of the rational function.

Example 1 :

r(x) = (3x⁵ + x⁴ - 6x³ - 2)/(x² - 9) 

Solution :

Equate the denominator to zero.

x² - 9 = 0

x² - 3² = 0

Using algebraic identity a² - b² = (a + b)(a - b),

(x + 3)(x - 3) = 0

x + 3 = 0   or   x - 3 = 0

x = -3   or   3

The domain is all real values except -3 and 3.

In interval notation,

(-∞, -3)U(-3, 3)U(3, +∞)

Example 2 :

r(x) = (5x³ − 12x² + 13)/(x² −  7)

Solution :

Equate the denominator to zero.

x² - 7 = 0

x² - (√7)² = 0

Using algebraic identity a² - b² = (a + b)(a - b),

(x + √7)(x - √7) = 0

x + √7 = 0   or   x - √7 = 0

x = -√7   or   √7

The domain is all real values except -√7 and √7.

In interval notation,

(-∞, -√7)U(-√7, √7)U(√7, +∞)

Example 3 :

r(x) = (x⁵ + 3x⁴ - 6)/(2x² − 5)

Solution :

Equate the denominator to zero.

2x² - 5 = 0

x² = 5/2

x = ±√(5/2)

The domain is all real values except ±√(5/2).

In interval notation,

(-∞, -√5/2)U(-√5/2,  √5/2)U(√5/2, +∞)

Example 4 :

r(x) = (4x⁷ + 8x² - 1)/(x² - 2x − 6)

Solution :

Equate the denominator to zero.

x² - 2x - 6 = 0

The above quadratic equation can not be solved by factoring.

Use quadratic formula to solve the equation.

Comparing ax² + bx + c = 0 and x² - 2x - 6 = 0,

a = 1, b = -2 and c = -6

Quadratic formula :

Substitute a = 1, b = -2 and c = -6.

The domain is all real values except 1 ± √7.

In interval notation,

(-∞, 1 - √7)U(1 - √7,  1 + √7)U(1 + √7, +∞)

Example 5 :

For each function, find the following :

i) determine the domain, intercepts, asymptotes, and positive/negative intervals

ii) use these characteristics to sketch the graph of the function

iii) describe where the function is increasing or decreasing

a) 

 f(x) = 2/(x - 3)

Solution :

a) Finding domain, intercepts, asymptotes and positive/negative intervals : 

By equating the denominator to 0, we get

x - 3 = 0

x = 3

So, domain is all real values except 3. That is (-∞, 3) and (3, ∞)

x-intercept :

Put y = 0

0 = 2/(x - 3)

So, there is no x-intercept.

y-intercept :

Put x = 0

y = 2/(0 - 3)

y = -2/3

Positive/negative intervals :

• When x = 0 ∈ (-∞, 3), f(x) is negative.
• When x = 4 ∈ (3, ∞), f(x) is positive.

Vertical asymptote :

x = 3

From the graph, the function is decreasing on its entire domain: when x ∈ (-∞, 3) and when x ∈ (3, ∞)

Example 6 :

Graph g(x) = [−4/(x + 2)] − 1. State the domain and range.

Solution :

g(x) = [−4/(x + 2)]  − 1

Vertical asymptote :

x + 2 = 0

x = -2

Horizontal asymptote :

Since the highest exponent of the numerator is equal to the highest exponent of the denominator, the horizontal asymptote will be calculated by coefficient of the numerator / coefficient of the denominator

g(x) = [−4 - (x + 2)]/(x + 2)

= (-4 - x - 2)/(x + 2)

= (-x - 6)/(x + 2)

Horizontal asymptote y = -1/1, then y = -1

Plot points to the left of the vertical asymptote, such as (−3, 3), (−4, 1), and (−6, 0). Plot points to the right of the vertical asymptote, such as (−1, −5), (0, −3), and (2, −2).

Draw the two branches of the hyperbola so that they pass through the plotted points and approach the asymptotes. The domain is all real numbers except −2 and the range is all real numbers except −1.

Example 7 :

Graph the function. State the domain and range.

f(x) = (x + 4)/(x - 3)

Solution :

f(x) = (x + 4)/(x - 3)

Vertical asymptote :

x - 3 = 0

x = 3

Horizontal asymptote :

y = 1/1

y = 1

Plotting the points (2, -6) (1, -2.5) (4, 8) (5, 4.5)

Domain is (-∞, 3) and (3, ∞)

Range is all real numbers except 1.

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