Question

For the following exercises, determine the point(s), if any, at which each function is discontinuous. Classify any discontinuity as jump, removable, infinite, or other. a) f(x)=xx2−x b) f(x)=x+3x2+5x+6 c) f(x)=|x−2|x−2

Answer 1

Answer:

a) Discontinuity is described to be infinite @ x = 1, and removable discontinuity @ x = 0

b) Discontinuity is described to be infinite @ x = -2, and removable discontinuity @ x = -3

c) Jump discontinuity @ x = 2

Step-by-step explanation:

a) f (x) = x / (x^2 - x)

To find discontinuities we need to see at what values of x is a function is either undetermined ( -∞ or ∞ ) or their is a jump within a function for certain values of x where it behaves differently:

In this case we have rational function. The most common discontinuity associated with rational functions is it being undetermined ( -∞ or ∞ ) at certain values of x. This can be found by simply equating the denominator equal to zero as follows:

f(x) = x / x.(x - 1) = 1 / (x-1)

We can see that original function f(x) was undetermined (0 / 0) at x = 0; however, after simplification the discontinuity was removed . So , a removable discontinuity @ x = 0:

denominator (x - 1 ) = 0

x = 1

Hence, f(1) = 1 / (1-1) = 1 / 0 = ∞  (Hence, discontinuity is described to be infinite @ x = 1)

Answer: Discontinuity is described to be infinite @ x = 1

b) f (x) = (x+3) / (3x^2 +5x + 6)

Similarly, in this we have a rational function again:

f (x) =  (x+3) / [(x+3)*(x+2)] =  1 / (x+2)

We can see that original function f(x) was undetermined (0 / 0) at x = -3; however, after simplification the discontinuity was removed . So , a removable discontinuity @ x = -3:

denominator (x + 2) = 0

x= -2

Hence, f(-2) = 1 / (-2+2) = 1 / 0 = ∞  (Hence, discontinuity is described to be infinite @ x = -2).

Answer: Discontinuity is described to be infinite @ x = -2

c) f(x) = |x - 2| / (x - 2)

In this case we have a disguised rational function to reveal its true nature we will uncloak the modulus from numerator first:

Hence,

g ( x ) = +(x-2) / (x - 2) = +1 for x > 2

h ( x ) = -(x-2) / (x-2) = -1  for x < 2

What we have is a disguised piece-wise function in which f(x) has two forms for a pair of x domain.

We can also see that for x = 2, f(x) is equal to both 1 and -1 as expressed as g(x) and h(x). This type of discontinuity is known to be jump discontinuity @ x = 2, where f(x) is step-function.

Answer: Jump discontinuity @ x = 2

Answer 2

Answer: (a). at x = 0, its a removable discontinuity

and at x = 1, it is a jump discontinuity

(b). at x = -3, it is removable discontinuity

also at x = -2, it is an infinite discontinuity

(c). at x = 2, it is a jump discontinuity

Step-by-step explanation:

in this question, we would analyze the 3 options to determine which points gave us discontinuous in the category of discontinuity as jump, removable, infinite, etc.

(a). given that f(x) = x/x² -x

this shows a discontinuous function, because we can see that the denominator equals zero i.e.

x² - x = 0

x(x-1) = 0

where x = 0 or x = 1.

since x = 0 and x = 1, f(x) is a discontinuous function.

let us analyze the function once more we have that

f(x) = x/x²-x = x/x(x-1) = 1/x-1

from 1/x-1 we have that x = 1 which shows a Jump discontinuity

also x = 0, this also shows a removable discontinuity.

(b). we have that f(x) = x+3 / x² +5x + 6

we simplify as

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

where x = -3, and x = -2 shows it is discontinuous.

from f(x) = x + 3 / (x + 3)(x + 2) = 1/x+2

x = -3 is a removable discontinuity

also x = -2 is an infinite  discontinuity

(c). given that f(x) = │x -2│/ x - 2

from basic knowledge in modulus of a function,

│x│= │x       x ˃ 0 and at │-x    x ∠ 0

therefore, │x - 2│= at │x - 2,     x ˃ 0 and at  │-(x - 2)   x ∠ 2

so the function f(x) = at│ 1,     x ˃ 2 and at │-1,    x ∠ 2

∴ at x = 2 , the we have a Jump discontinuity.

NB. the figure uploaded below is a diagrammatic sketch of each of the function in the question.

cheers i hope this helps.