Question

Which of the following functions f has a removable discontinuity at a? If the discontinuity is removable, find a function g that agrees with f for x ≠ a and is continuous at a. (If an answer does not exist, enter DNE.)
(a) f(x) = x4 − 1 x − 1 , a = 1 g(x) =
(b) f(x) = x3 − x2 − 56x x − 8 , a = 8 g(x) =
(c) f(x) = sin x , a = π (Recall that h(x) means the largest integer that is less than or equal to h(x).) g(x) =

Answer 1

Answer:You can remove the discontinuity at

x = 3 by defining g(3) = –1.

Answer 2

Using the continuity concept, it is found that the correct options are given by options a and b, which are:

$f(x) = \frac{x^4 - 1}{x - 1}$

$f(x) = \frac{x^3 - x^2 - 56x}{x - 8}$

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• A function is not continuous at a value of x which is outside the domain.
• If the function can be factored at this point, the discontinuity is removable, otherwise, it is non-removable.

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Item a:

The function is given by:

$f(x) = \frac{x^4 - 1}{x - 1}$

• A test is made is a = 1.
• 1 is a root of the denominator, thus, we have to verify if it is at the numerator. If it is, the discontinuity is removable.

$(1)^4 - 1 = 1 - 1 = 0$

• It is a root at the numerator, thus, the discontinuity is removable.

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Item b:

The function is given by:

$f(x) = \frac{x^3 - x^2 - 56x}{x - 8}$

• A test is made is a = 8.
• 8 is a root of the denominator, thus, we have to verify if it is at the numerator. If it is, the discontinuity is removable.

$8^3 - 8^2 - 56(8) = 0$

• It is a root at the numerator, thus, the discontinuity is removable.

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Item c:

The function is:

$f(x) = \sin{x}$

• All values of x are in the domain, so it has no discontinuities.

A similar problem is given at https://brainly.com/question/12701640