Examining Discontinuity- Problem # 2

[Jason76]

Post Edited

\(\displaystyle f(x) = \dfrac{x - 3}{x^{2} - x - 6}\)

a. Find all discontinuities

Finding the asymptotic ones would involve finding what values of x would cause a 0 in the denominator (undefined).

b. Classify the discontinuities as either removable or asymptotic.

Finding the asymptotic ones would involve finding what values of x would cause a 0 in the denominator (undefined), but what about finding the removable ones?

 

[Jason76]

First of all you would need to factor the denominator out. If a term in the denominator matches one in the numerator, then it cancels out. Set the matching terms to 0 and solve for x. Whatever x equals is the removable discontinuity. However, regarding the term that doesn't cancel out. It is the asymptotic discontinuity. Right

 

[stapel]

\(\displaystyle f(x) = \dfrac{x - 3}{x^{2} - x - 6}\)

a. Find all discontinuities

Finding the asymptotic ones would involve finding what values of x would cause a 0 in the denominator (undefined).

Yes. So apply what you learned back in algebra.

b. Classify the discontinuities as either removable or asymptotic.

Finding the asymptotic ones would involve finding what values of x would cause a 0 in the denominator (undefined), but what about finding the removable ones?

The "removable" discontinuities are the ones that were "the special case with a hole" back in algebra.