Question Video: Discussing the Continuity of a Rational Function over Its Domain | Nagwa Question Video: Discussing the Continuity of a Rational Function over Its Domain | Nagwa

Question Video: Discussing the Continuity of a Rational Function over Its Domain Mathematics • Second Year of Secondary School

Join Nagwa Classes

Attend live General Mathematics sessions on Nagwa Classes to learn more about this topic from an expert teacher!

Discuss the continuity of the function 𝑓 given 𝑓(π‘₯) = (π‘₯Β² βˆ’ 4)/(π‘₯ + 2). [A] The function is continuous on ℝ βˆ’ {βˆ’2}. [B] the function is continuous on ℝ.

03:02

Video Transcript

Discuss the continuity of the function 𝑓 given 𝑓of π‘₯ equals π‘₯ squared minus four over π‘₯ plus two. The two options are A) the function is continuous on the set of real numbers ℝ minus the set of negative two and B) the function is continuous on the set of real numbers ℝ.

The only difference between these two options is whether negative two is included in the set on which 𝑓 is continuous. So is 𝑓 continuous at π‘₯ equals negative two? Recall that a function 𝑓 is said to be continuous at a number 𝑐 if the limit of 𝑓 of π‘₯ as π‘₯ approaches 𝑐 is just 𝑓 of 𝑐.

And the somewhat hidden requirement here is that the quantities on both sides of this equation must exist. So the limit of 𝑓 of π‘₯ as π‘₯ approaches 𝑐 must exist and 𝑓 of 𝑐 must exist. So 𝑐 must be in the domain of 𝑓.

Looking at the rule of 𝑓 of π‘₯, we can see that negative two isn’t in the domain of 𝑓. If we try to use the rule to evaluate 𝑓 of negative two, we get the indeterminate form zero over zero. And as negative two isn’t in the domain of 𝑓, 𝑓 can’t possibly be continuous at negative two. Negative two is therefore not in the set on which 𝑓 is continuous. Our answer is therefore A the function is continuous on the set of real numbers with negative two removed.

We can check if we’d like to that 𝑓 is indeed continuous at all of the real numbers apart from negative two. But we don’t need to to decide between the two options we have. To do this, we can use the fact that a rational function is continuous on its domain.

This is something that we can prove using the laws of limits and the domain of our rational function is the set of real numbers minus the single real number, which makes the denominator zero. That is negative two.

You might have been tempted to factor the numerator of this rational function, which is the difference of two squares, and then to cancel the common factor of π‘₯ plus two, leaving just π‘₯ minus two. If we let 𝑔 of π‘₯ equals π‘₯ minus two, then 𝑓 of π‘₯ is equal to 𝑔 of π‘₯ for any value of π‘₯ apart from negative two and 𝑔 of negative two is negative four.

If you had simplified 𝑓 of π‘₯ two π‘₯ minus two, then you might think that 𝑓 of negative two was negative four as well. And then you would probably think that option B was the right answer and that 𝑓 of π‘₯ was continuous on the whole set of real numbers ℝ. But unfortunately, this isn’t the case. 𝑓 of negative two not really is undefined. Even though simplifying it somehow, you’d think that the value should be negative four.

We can however use 𝑔 of π‘₯ to show that the limit of 𝑓 of π‘₯ as π‘₯ approaches negative two is negative four. While 𝑓 of negative two is not equal to 𝑔 of negative two, the limit of 𝑓 of π‘₯ as π‘₯ approaches negative two is equal to the limit of 𝑔 of π‘₯ as π‘₯ approaches negative two.

Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

  • Interactive Sessions
  • Chat & Messaging
  • Realistic Exam Questions

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy