How would you "remove the discontinuity" of f? In other words, how would you define

f (5) in order to make f continuous at 5?

f (x) = [ (x^2 - 2x - 15) / x-5]

Question

Eric Skinner

Mathematics

25 January, 12:24

How would you "remove the discontinuity" of f? In other words, how would you define

f (5) in order to make f continuous at 5?

f (x) = [ (x^2 - 2x - 15) / x-5]

+5

Answers (1)

Know the Answer?

Abram Booker · Answer 1

The numerator of this function given may be factored out as follows,

x² - 2x - 15 = (x - 5) (x + 3)

Given that the expression x - 5 appears in both the numerator and the denominator, we can cancel them and the new function becomes,

f (x) = x + 3

which is continuous at 5.

How would you "remove the discontinuity" of f? In other words, how would you definef (5) in order to make f continuous at 5?f (x) = [ (x^2 - 2x - 15) / x-5]

How would you "remove the discontinuity" of f? In other words, how would you define

f (5) in order to make f continuous at 5?

f (x) = [ (x^2 - 2x - 15) / x-5]