Video: The Intermediate Value Theorem

The function 𝐹(π‘₯) = 1/π‘₯ + 3 satisfies 𝐹(βˆ’1) < 3 and 𝐹(1) > 3. But there is no π‘₯ between βˆ’1 and 1, where 𝐹(π‘₯) = 3. Why does this not violate the intermediate value theorem? [A] because the function 𝐹 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 𝐹(π‘₯) = 0, not 𝐹(π‘₯) = 3 [D] because the function is not defined on the entire interval [βˆ’1, 1] [E] becau

04:44

Video Transcript

The function 𝐹 of π‘₯ equals one over π‘₯ plus three satisfies 𝐹 of negative one is less than three and 𝐹 of one is greater than three. But there is no π‘₯ between negative one and one, where 𝐹 of π‘₯ equals three. Why does this not violate the intermediate value theorem? Is it A) because the function 𝐹 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 𝐹 of π‘₯ is zero, not 𝐹 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.

Let’s clear the options from the screen so we can focus on the question. Let’s have a look at the statement of the intermediate value theorem. If 𝑓 is continuous on the closed interval from π‘Ž to 𝑏 and some number 𝑛 is between 𝑓 of π‘Ž and 𝑓 of 𝑏, then there exists some number 𝑐 in the open interval from π‘Ž to 𝑏 with 𝑓 of 𝑐 equal to 𝑛.

Looking at our function, we’re told that 𝐹 of negative one is less than three and 𝐹 of one is greater than three. Let’s check that. To find 𝐹 of negative one, we substitute negative one for π‘₯ in the definition of 𝐹 of π‘₯ and then evaluate. So we get one over negative one plus three, which is two. And it’s a similar process to find 𝐹 of one, which is four. And indeed, two is less than three and four is greater than three.

We’re also told that there is no π‘₯ between negative one and one, where 𝐹 of π‘₯ is equal to three. We check this by trying to solve the equation 𝐹 of π‘₯ equals three. We use the definition of 𝐹 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 𝐹 of π‘₯ is equal to three.

Why is this a problem? Well, look at the statement of the intermediate value theorem. We’re told that if 𝑛 is between 𝑓 of π‘Ž and 𝑓 of 𝑏 and we’ve seen in our example that three is between 𝐹 of negative one and 𝐹 of one, then there should exist 𝑐 in the open interval from π‘Ž to 𝑏 β€” in our case from negative one to one β€” with 𝑓 of 𝑐 equal to 𝑛 β€” in our case three.

But we’ve just seen that there is no 𝑐 in the open interval from negative one to one with 𝑓 of 𝑐 equal to three. So what’s gone wrong? Is the intermediate value theorem incorrect? Have we found the counterexample? Well, the theorem is correct. But it has a condition 𝑓 must be continuous on the closed interval π‘Ž, 𝑏.

So our function capital 𝐹 must be continuous on the closed interval negative one to one. And it isn’t continuous on this interval. And that’s why the intermediate value theorem doesn’t hold. Let’s show that our function 𝑓 isn’t 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 𝑓 is continuous on the closed interval from π‘Ž to 𝑏. If all values 𝑐 in the open interval from π‘Ž to 𝑏, 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 𝑓 of π‘₯ as π‘₯ approaches 𝑐 must exist for all numbers 𝑐 in the open interval and 𝑓 of 𝑐 must exist for all values in the open interval. For our function capital 𝐹, zero is in the closed interval from negative one to one. But if you try to evaluate 𝐹 of zero, we get one over zero plus three, which is undefined.

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

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

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