This prompt is about removable discontinuities. See the explanation below.
A removable discontinuity is a point in a graph where it is not linked but may be made so by filling in a single point.
It is also possible to define it as follows:
A discontinuity is detachable at x=a if the limit limxaf(x) exists and is finite. There are two kinds of removable discontinuities. At x=a, the function is undefined.
It should be noted that a non-removable discontinuity is one in which the limit of the function does not exist at a given point, i.e. lim xa f(x) does not exist.
Part A: Where F(x) = 6/x
At x = 0
f(x) = 6/0 = ∞; Thus,
At x = 0
f(x) is not defined. It is correct to state therefore, that f is continuous for all real integers or number save "zero".
Hence, f(x) is continuous at each x ∈ R - {α}
Part B: Where F(x) = 4/(x-6)
At x = 6
F(x) = 4/(6-6)
= 4/0
= ∞;
Thus, f (x) is not defined.
We can state therefore that F is continuous at x ∈ R - { α}
Part C: Where F (x)
F(x) = x² - 9
For each C∈R,
F(c) = C² = 9
Thus, F(x) here is defined and continuous. That is F(x) is continuous at x ∈ R
Part D: Where F(x) x² - 4x + 4
With respect to every C ∈ R,
F(c) = C² - 4c + 4
In this instance as well, F(c) is defined and continuous.
Thus, F(x) in this case is continuous for all X ∈ R
Learn more about removable discontinuity:
https://brainly.com/question/23655932
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Full Question:
Removable and nonremovable discontinuities. In exercises 35–60, find the x-values (if any) at which f is not continuous. Which of the discontinuities are removable?
35: f(x) = 6/x
36: f(x) = 4/(x-6)
37: f(x) = x² - 9
38: f(x) x² - 4x + 4
