HOW TO IDENTIFY POINTS OF DISCONTINUITY

Find and classify any points of discontinuity 

Example 1 :

y = (x² + x - 6) / (3x² - 12)

Solution :

y = (x² + x - 6) / (3x² - 12)

Factoring the numerator and denominator, we get

x² + x - 6 = x² + 3x - 2x - 6

= x( x + 3) - 2(x + 3)

= (x + 3)(x - 2)

3x² - 12 = 3(x² - 4)

= 3(x² - 2²)

= 3(x + 2)(x - 2)

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

The common factors is x - 2, using zero property

x - 2 = 0

x = 2

At x = 2, the function will have a hole.

x + 2 = 0

x = -2

At x = -2, the function has a vertical asymptote.

Example 2 :

y = x/(x² - 9)

Solution :

y = x/(x² - 9)

Factoring the numerator and denominator, we get

x² - 9 = (x² - 3²)

= (x + 3)(x - 3)

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

There is no common factor in both numerator and denominator. 

x + 3 = 0 and x - 3 = 0

x = -3 and x = 3

At x = -3 and x = 3, the function has vertical asymptotes.

Example 3 :

y = (3x² - 1)/x³

Solution :

y = (3x² - 1)/x³

There is no common factor in both numerator and denominator. 

x = 0

At x = 0, the function has vertical asymptotes.

Example 4 :

Find constants a and b such that the function

continuous for -3 ≤ x ≤ 0.

Solution :

lim _x->-2-f(x) = lim _x->-2 -x

= -(-2)

= 2

lim _x->-2+ f(x) = lim _x->-2+ ax² + b

= a(-2)² + b

= 4a + b

lim _x->-2-f(x) = lim _x->-2+ f(x)

4a + b = 2 -----(1)

lim _x->0- f(x) = lim _x->0- ax² + b

= a(0)² + b

= b

lim _x=0 f(x) = lim _x=0 6

= 6

Since the function f(x) is continuous between -3 and 0,

lim _x->-2-f(x) = lim _x->-2+ f(x) = lim _{x = -2} f(x)

lim _x->0-f(x) = lim _x->0+ f(x) = lim _{x = 0} f(x)

b = 6

Applying the value of b in (1), we get

4a + b = 2

4a + 6 = 2

4a = 2 - 6

4a = -4

a = -4/4

a = -1

So, the values of a and b are -1 and 6 respectively.

Example 5 :

Find constants a and b such that the function

is continuous.

Solution :

lim _x->3-f(x) = lim _x->3- x² - 1

= 3² - 1

= 9 - 1

= 8

lim _x->3+ f(x) = lim _x->3⁺ 2ax

= 2a(3)

= 6a

Since the function is continuous, 

lim _x->3-f(x) = lim _x->3+ f(x)

8 = 6a

a = 8/6

a = 4/3

So, the value of a is 4/3.

Example 6 :

Examine the continuity of g(x) = x + 3 at the point x = 2.

Solution :

g(x) = x + 3 at the point x = 2

Since it is continuous function, then

g(2) = 2 + 3

g(2) = 5

Find the value(s) of x at which the functions are discontinuous.

Example 7 :

f(x) = (9 - x²) / (x - 3)

Solution :

f(x) = (9 - x²) / (x - 3)

f(x) = (3² - x²) / (x - 3)

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

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

= -(3 + x)

Equating the common factor to 0, we get

x - 3 = 0

x = 3

The function is discontinuous at x = 3

Example 8 :

f(x) = 13x / (x² + x - 6)

Solution :

f(x) = 13x / (x² + x - 6)

x² + x - 6 = x² + 3x - 2x - 6

= x(x + 3) - 2(x + 3)

= (x - 2)(x + 3)

f(x) = 13x / (x - 2)(x + 3)

x - 2 = 0 and x + 3 = 0

x = 2 and x = -3

So, the function is discontinuous at x = 2 and x = -3.

Example 9 :

Solution :

lim _x->3- f(x) = lim _x->3- -x

= -3

lim _x->3+ f(x) = lim _x->3+ (1 - x)

= 1 - 3

= -2

From the above proofs, 

lim _x->3- f(x) = lim _x->3+ f(x)

The function f(x) is discontinuous.

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