Video: Using the Intermediate Value Theorem to Determine the Properties of a Continuous Function

Consider the continuous function 𝑓(π‘₯). Given that 𝑓(5) = βˆ’2 and 𝑓(8) = 4, which of the following must be true according to the intermediate value theorem? [A] There is a least one zero on the interval [5, 8]. [B] There is exactly one zero on the interval [5, 8]. [C] There are no zeros on the interval [5, 8]. [D] There are no zeros on the intervals (βˆ’βˆž, 5] and [8, ∞). [E] The only zeros of 𝑓(π‘₯) lie on the interval [5, 8].

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Video Transcript

Consider the continuous function 𝑓 of π‘₯. Given that 𝑓 evaluated at five is equal to negative two and 𝑓 evaluated at eight is equal to four, which of the following must be true according to the intermediate value theorem? Option (A) there is a least one zero on the closed interval from five to eight. Option (B) there is exactly one zero on the closed interval from five to eight. Option (C) there are no zeros on the closed interval from five to eight. Option (D) there are no zeros on the intervals where π‘₯ is less than or equal to five and π‘₯ is greater than or equal to eight. Or option (E) the only zeros of 𝑓 of π‘₯ lie on the closed interval from five to eight.

The question gives us a function 𝑓 of π‘₯ which we’re told is continuous. We’re told that 𝑓 evaluated at five is equal to negative two and 𝑓 evaluated at eight is equal to four. We need to use this information and the intermediate value theorem to decide which of our statements must be true. Let’s start by recalling what the intermediate value theorem tells us. The intermediate value theorem says if 𝑓 is a continuous function on a closed interval from π‘Ž to 𝑏 and we have a constant 𝑁 which is between 𝑓 evaluated at π‘Ž and 𝑓 evaluated at 𝑏. Then the intermediate value theorem says there must exist a 𝑐 on the open interval from π‘Ž to 𝑏 where 𝑓 evaluated at 𝑐 is equal to 𝑁.

So the concluding statement of the intermediate value theorem tells us about the existence of one value of 𝑐 where 𝑓 of 𝑐 is equal to 𝑁. And if we look at our answers, all of our answers are talking about the number of zeros our function 𝑓 of π‘₯ has, all the number of zeros it has on some specific interval. So, it makes sense to set our value of 𝑁 equal to zero. Then, the concluding statement of the intermediate value theorem tells us about the existence of a zero of our function 𝑓 of π‘₯. But remember, the question also tells us 𝑓 evaluated at five is equal to negative two, and 𝑓 evaluated at eight is equal to four. We can see that zero is between negative two and four. So if we set our value of π‘Ž equal to five and our value of 𝑏 equal to eight, we can use the intermediate value theorem.

Let’s quickly check the prerequisites of the intermediate value theorem. First, in the question, we’re told our function 𝑓 of π‘₯ is continuous. This means it’s continuous for all real values of π‘₯. So, in particular, it will be continuous on any closed interval. Next, we need to check that our value of 𝑁 is between 𝑓 evaluated at π‘Ž and 𝑓 evaluated at 𝑏. To start, our value of π‘Ž is five, and we know 𝑓 evaluated at five is negative two. This is less than zero. And we chose 𝑏 equal to eight, and we know 𝑓 evaluated at eight is equal to four. This is bigger than zero. So 𝑁 is between 𝑓 evaluated at π‘Ž and 𝑓 evaluated at 𝑏. So all the prerequisites for our intermediate value theorem are true. So we can conclude that there must exist a 𝑐 in the open interval from five to eight where 𝑓 evaluated at 𝑐 is equal to zero.

So let’s check this against each of our options. Option (A) says that there should be at least one zero in the closed interval from five to eight. Well, if we’ve proven there’s at least one zero in the open interval from five to eight, there must also be at least one zero in the closed interval from five to eight. So option (A) is true. However, we don’t know whether option (B) is true. There might be one zero but there might be more. Option (C) tells us there are no zeros for our function in the closed interval from five to eight. However, the intermediate value theorem told us there must be at least one zero in the open interval from five to eight. Since this is contained in the closed interval from five to eight, option (C) can’t be true.

Option (D) tells us about the number of zeros when π‘₯ is less than or equal to five or π‘₯ is greater than or equal to eight. However, we were only able to include information about π‘₯ in the open interval from five to eight. These values of π‘₯ are not included in our two intervals, so we can’t make any decision about option (D). And the same goes for option (E). We were only able to show by using the intermediate value theorem, there exists at least one 𝑐 in the open interval where 𝑓 of 𝑐 is equal to zero. We don’t know what happens outside of this open interval, so the only option we can conclude is option (A). We could end here, but we can also show that our remaining options must also be false.

We’ll show that options (B), (D), and (E) can’t be true by constructing possible functions 𝑓 of π‘₯ where these statements are not true. Let’s start with option (B). We want to find a function 𝑓 of π‘₯ which is continuous, 𝑓 evaluated at five is negative two, 𝑓 evaluated at eight is equal to four, and it has more than one zero on the closed interval from five to eight. To start, since 𝑓 of five must be equal to negative two and 𝑓 of eight must be equal to four, our function must pass through the two coordinates. Next, we want our function to have more than one zero on our closed interval from five to eight. And now we can see multiple ways of doing this. We’ll do this by using a piecewise-defined function.

The idea behind our piecewise-defined function is we start at the coordinates five, negative two and draw a straight line up to the π‘₯-axis where π‘₯ is equal to six. Then, we follow along the π‘₯-axis from π‘₯ is equal to six to π‘₯ is equal to seven. This means that our function will be equal to zero for all values of π‘₯ between six and seven. So 𝑓 of π‘₯ has multiple zeros on this interval. Then again, we just draw the straight line from seven up to our other coordinate, eight, four. And it doesn’t really matter what our function does below five and above eight as long as it’s continuous. So we just made these equal to negative two and four, respectively.

Now, by using this example, we can see that there is a function 𝑓 of π‘₯ which has more than one zero on the closed interval from five to eight which satisfies these properties. To talk about options (D) and (E), we’re going to modify our piecewise-defined function slightly. Now, when π‘₯ is less than five, instead of being the constant negative two, we’re going to connect it up to the point in the π‘₯-axis where π‘₯ is equal to three. Our function is still continuous and passes through both of our points. However, we now have a new zero when π‘₯ is equal to three. We can see this means option (D) can’t be true since 𝑓 of three is equal to zero. And option (E) also can’t be true since we now have a zero outside of this closed interval.

Therefore, we were able to show that if a continuous function 𝑓 of π‘₯ has 𝑓 evaluated at five is equal to negative two and 𝑓 evaluated at eight is equal to four. Then, the intermediate value theorem tells us that 𝑓 of π‘₯ has at least one zero on the closed interval from five to eight.

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