TYPES OF DISCONTINUITY ALGEBRAICALLY

What is continuous ?

Functions that can be drawn without lifting up your pencil are called continuous functions. You will define continuous in a more mathematically rigorous way after you study limits. 

Types of discontinuity :

There are three types of discontinuities:

i) Removable,

ii) Jump and

iii) Infinite

Discuss the continuity. If a discontinuity exists, then describe the type of discontinuity and its physical feature on a graph.

Problem 1 :

Solution :

Removable discontinuity :

Common factor is (x - 3), by equating x - 3 to 0, we get x = 3. So, removable discontinuity at x = 3 or hole is at x = 3.

Non removable discontinuity :

We find vertical asymptote, we equate the denominator to 0. So, we get x = 1.

Problem 2 :

Solution :

So, non removable discontinuity is at x = 3 or jump discontinuity is at x = 3.

Problem 3 :

Solution :

So, non removable discontinuity is at x = 3 or jump discontinuity is at x = 3.

Problem 4 :

Solution :

Removable discontinuity is at x = 0 or hole is at x = 0.

Problem 5 :

Solution :

Removable discontinuity is at x = 1 or hole is at x = 1.

Problem 6 :

Solution :

lim x --> -4 f(x) = -6

lim x -->-4⁺ f(x) = lim x -->-4^- f(x) = lim x -->-4 f(x)

The function is continuous everywhere.

Problem 7 :

For which of the following does lim x--> 4 exists ?

a) I only     b) II only     c) III only     d) I and II only      e)  I and III only

Solution :

By observing graphs I and II, when lim x--> 4 exists. In graph III, there is jump discontinuity. 

lim x--> 4^- f(x) = 4

lim x--> 4⁺ f(x) = 2

Since the limits are not equal, then the limit does not exists.

Problem 8 :

For the function identify the type of each discontinuity and where it is located.

f(x) = (x² - 8x + 12)/(x² + 3x - 10)

Solution :

f(x) = (x² - 8x + 12)/(x² + 3x - 10)

Factoring the numerator and the denominator, we get

f(x) = (x - 6)(x - 2)/(x + 5)(x - 2)

Since we see the common factors in both numerator and denominator, there is removable discontinuity at x = 2. Discontinuous at x = -5.

Problem 9 :

Determine if each function is continuous. If the function is not continuous, find the x-axis location of and classify each discontinuity.

f(x) = -x³ + x² - 3

Solution :

f(x) = -x³ + x² - 3

Since it is polynomial function, it is continuous everywhere.

Problem 10 :

f(x) = -(x + 2) / (x² - 4)

Solution :

f(x) = -(x + 2) / (x² - 4)

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

Since the common factor is x + 2, there is removable discontinuity at x = -2 and we have infinite discontinuity at x = 2.

Problem 11 :

Solution :

lim x--> -3^- f(x) = lim x--> -3^- (-x² - 6x - 8)

= (-(-3)² - 6(-3) - 8)

= -9 + 18 - 8

= -17 + 18

= 1

lim x--> -3⁺ f(x) = lim x--> -3⁺ (x - 2)

= -3 - 2

= -5

Since it is piecewise function and the limits are not same, there is a jump discontinuity.

Problem 12 :

The graph of the function 𝑓(𝑥) is shown to the right: Which of the following statements is true about 𝑓?

I. 𝑓 is undefined at 𝑥 = 1.

II. 𝑓 is defined but not continuous at 𝑥 = 2.

III. 𝑓 is defined and continuous at 𝑥 = 3.

(A) Only I     (B) Only II     (C) I and II     (D) I and III

(E) None of the statements are true.

Solution :

I. 𝑓 is undefined at 𝑥 = 1.

When x = 1, y = 3. So, it is false.

II. 𝑓 is defined but not continuous at 𝑥 = 2.

It is continuous at x = 2, because to the left and right of 2 we have continuity of the graph. It is false.

III. 𝑓 is defined and continuous at 𝑥 = 3.

At x = 3, we have infinite discontinuity. It is false.