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.
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
