### Video Transcript

The function πΉ of π₯ equals one over π₯ plus three satisfies πΉ of negative one is less than three and πΉ of one is greater than three. But there is no π₯ between negative one and one, where πΉ of π₯ equals three. Why does this not violate the intermediate value theorem? Is it A) because the function πΉ is not continuous over its domain, B) because the intermediate value theorem only applies to polynomial functions, C) because the intermediate value theorem only applies to cases, where πΉ of π₯ is zero, not πΉ of π₯ is three, D) because the function is not defined on the entire interval from negative one to one, or E) because the intermediate value theorem only applies on the open interval from zero to infinity.

Letβs clear the options from the screen so we can focus on the question. Letβs have a look at the statement of the intermediate value theorem. If π is continuous on the closed interval from π to π and some number π is between π of π and π of π, then there exists some number π in the open interval from π to π with π of π equal to π.

Looking at our function, weβre told that πΉ of negative one is less than three and πΉ of one is greater than three. Letβs check that. To find πΉ of negative one, we substitute negative one for π₯ in the definition of πΉ of π₯ and then evaluate. So we get one over negative one plus three, which is two. And itβs a similar process to find πΉ of one, which is four. And indeed, two is less than three and four is greater than three.

Weβre also told that there is no π₯ between negative one and one, where πΉ of π₯ is equal to three. We check this by trying to solve the equation πΉ of π₯ equals three. We use the definition of πΉ of π₯, subtract three from both sides, and rearrange to get π₯ equals one over zero, which is undefined. So in fact, there is no value of π₯ for which πΉ of π₯ is equal to three.

Why is this a problem? Well, look at the statement of the intermediate value theorem. Weβre told that if π is between π of π and π of π and weβve seen in our example that three is between πΉ of negative one and πΉ of one, then there should exist π in the open interval from π to π β in our case from negative one to one β with π of π equal to π β in our case three.

But weβve just seen that there is no π in the open interval from negative one to one with π of π equal to three. So whatβs gone wrong? Is the intermediate value theorem incorrect? Have we found the counterexample? Well, the theorem is correct. But it has a condition π must be continuous on the closed interval π, π.

So our function capital πΉ must be continuous on the closed interval negative one to one. And it isnβt continuous on this interval. And thatβs why the intermediate value theorem doesnβt hold. Letβs show that our function π isnβt continuous on this interval. And therefore that the intermediate value theorem needs not hold.

Remember the definition of a function being continuous on an interval. The function π is continuous on the closed interval from π to π. If all values π in the open interval from π to π, the function is continuous at that point. And also the one-sided limit at each end point must agree with the value of the function there.

In particular, the limit of π of π₯ as π₯ approaches π must exist for all numbers π in the open interval and π of π must exist for all values in the open interval. For our function capital πΉ, zero is in the closed interval from negative one to one. But if you try to evaluate πΉ of zero, we get one over zero plus three, which is undefined.

As a result, πΉ of π is not defined for all values π in the open interval from negative one to one. That is it is defined for most of these values π but not all of them because πΉ of zero is undefined. And as a result, our function capital πΉ does not satisfy the definition to be continuous on the interval negative one, one. πΉ is not continuous on this interval and so the intermediate value theorem it doesnβt apply.

The intermediate value theorem is not violated because it doesnβt apply here. It doesnβt apply because πΉ is not continuous on the closed interval from negative one to one and is not continuous on this closed interval because the function πΉ is not defined on the entire interval.