Question Video: Applying the Intermediate Value Theorem Mathematics • Higher Education

The figure shows the graph of the function 𝑓 on the interval [0, 16] together with the dashed line 𝑦 = 30. 𝑓(0) < 30 and 𝑓(16) > 30, but 𝑓(π‘₯) β‰  30 anywhere on [0, 16]. Why does this not violate the intermediate value theorem?

02:04

Video Transcript

The figure shows the graph of the function 𝑓 on the closed interval from zero to 16 together with the dashed line with equation 𝑦 equals 30. 𝑓 of zero is less than 30 and 𝑓 of 16 is greater than 30. But 𝑓 of π‘₯ is not equal to 30 anywhere on the closed interval from zero to 16. Why does this not violate the intermediate value theorem?

Let’s just verify what we’re told in the question. Is 𝑓 of zero less than 30? Well, yes, we can see here 𝑓 of zero appears to be about 12. And likewise, 𝑓 of 16 is greater than 30. It appears to be about 32. But 𝑓 of π‘₯ is not equal to 30 anywhere on the interval. This is true because the dashed line 𝑦 equals 30 nowhere intersects the graph of our function. The question is, why does this not to violate the intermediate value theorem? Well, what is the intermediate value theorem? It states that if a function 𝑓 is continuous on a closed interval from π‘Ž to 𝑏 and if the number 𝑁 is between 𝑓 of π‘Ž and 𝑓 of 𝑏, the values of the function on the end points of the interval. Then there exists 𝑐 in the open interval from π‘Ž to 𝑏 such that 𝑓 of 𝑐 is 𝑁.

We can see then how it might appear that we have a counterexample to the intermediate value theorem. We set 𝑁 equal to 30 and note that 30 is between 𝑓 of zero and 𝑓 of 16. However, there doesn’t exist any value 𝑐 in the open interval from zero to 16 such that 𝑓 of 𝑐 is 30. Why is this not a counterexample to the intermediate value theorem? Well, the intermediate value theorem only applies if 𝑓 is continuous. Our function doesn’t satisfy this required hypothesis. We can see that there is a discontinuity here at π‘₯ equals eight.

So why does this not violate the intermediate value theorem? Because the function is not continuous at π‘₯ equals eight. And so, it is not continuous on the closed interval from zero to 16 which would be required for the intermediate value theorem to apply.

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