Answers

1. This prompt is about removable discontinuities. See the explanation below.

   What is a removable discontinuity?

   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.

   What is the calculation justifying the above answer?

   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:
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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

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