Which equation could be used to solve the question below?
2 answers:
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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.
