HOW TO FIND A POINT OF DISCONTINUITY OF RATIONAL FUNCTION

We say a function is continuous if its domain is an interval, and it is continuous at every point of that interval.

A point of discontinuity is the only bad point for the function on some interval.

How to find point of discontinuity of a rational function ?

The discontinuities of a rational function can be found by setting its denominator equal to zero and solving it.

The function given above is not continuous at x = 1

Types of discontinuity :

(i) Removable

(ii) Jump

(iii)  Essential

(iv) Infinite

Find any points of discontinuity for each rational function.

Problem 1 :

y = (x + 3)/(x – 4) (x + 3)

Solution :

To find points of discontinuity, let us equate the denominators to 0.

y = (x + 3)/(x – 4) (x + 3)

x – 4 = 0

x = 4

x + 3 = 0

x = -3

The function is discontinuous at x = -3 and 4.

Problem 2 :

y = (x - 2)/(x² – 4)

Solution :

To find point of discontinuity, let us equate the denominator to 0.

y = (x - 2)/(x² – 4)

x² – 4 = 0

x² = 4

x = ±2

The function is discontinuous at x = ±2.

Problem 3 :

y = (x - 3) (x + 1)/(x – 2)

Solution :

To find point of discontinuity, let us equate the denominator to 0.

y = (x - 3) (x + 1)/(x – 2)

x – 2 = 0

x = 2

The function is discontinuous at x = 2.

Problem 4 :

y = 3x(x + 2)/x(x + 2)

Solution :

To find point of discontinuity, let us equate the denominator to 0.

y = 3x(x + 2)/x(x + 2)

x(x + 2) = 0

x = 0, x + 2 = 0

x = 0 and x = -2

The function is discontinuous at x = 0, -2.

Problem 5 :

y = 2/(x + 1)

Solution :

To find point of discontinuity, let us equate the denominator to 0.

y = 2/(x + 1)

x + 1 = 0

x = -1

The function is discontinuous at x = -1.

Problem 6 :

y = 4x/(x³ – 9x)

Solution :

To find point of discontinuity, let us equate the denominator to 0.

y = 4x/(x³ – 9x)

x³ – 9x = 0

x(x² - 9) = 0

Equating each factor to zero, we get

x = 0, x² - 9 = 0

x=  0 and x = ±3

The function is discontinuous at x = 0, ±3.

Problem 7 :

Determine whether

f(x) = (x² - x - 6)/(x - 3)

is continuous at x = 3.

Solution :

f(x) = (x² - x - 6)/(x - 3)

Factoring the numerator, we get

= (x - 3)(x + 2)/(x - 3)

Since the common factor is (x - 3), there is removable discontinuity at x = 3. 

f(x) = x + 2

Problem 8 :

Find the points of discontinuity of the function and identify the type of discontinuity.

f(x) = (x² - 4x + 3)/(x - 1)

Solution :

f(x) = (x² - 4x + 3)/(x - 1)

= (x - 1)(x + 4)/(x - 1)

Since the common factor is (x - 1), there is removable discontinuity at x = 1. 

f(x) = x + 4

Problem 9 :

Find the value of k in each of the following:

is continuous at x = 5

Solution :

lim x--> 5^- f(x) =  lim x--> 5^- kx + 1

Applying x = 5, we get

= 5k + 1 -----(1)

lim x--> 5⁺f(x) =  lim x--> 5⁺ 3x - 5

Applying x = 5, we get

= 3(5) - 5

= 15 - 5

= 10 -----(2)

(1) = (2)

5k + 1 = 10

5k = 10 - 1

5k = 9

k = 9/5

Problem 10 :

Use the function f defined and graphed below to answer the questions.

a) Does f(-1) exists ?

b) Does lim x-> 1^+ f(x) exists ?

c)  Does lim x-> 1^+ f(x) = f(-1)

d) Is f continuous at x = -1

Solution :

a) When x = -1, the function should be f(x) = x^2 - 1

applying x = -1, we get

= (-1)^2 - 1

= 1 - 1

= 0

So, f(-1) = 0

b) lim x-> 1^+ f(x), since we have no break after x = 1, it is continuous,

c)  lim x-> 1^+ f(x) = -2(1) + 4

= 2

f(-1) = 0

They are not equal.

d) Since we have a filled circle at x = -1, it is continuous.