Question Video: Using the Intermediate Value Theorem to Determine Information about Functions Represetning a Particle′s Position Mathematics

Video Transcript

A particle moving along a line has at each time 𝑡 a position function 𝑠 of 𝑡, which is continuous. Assume that 𝑠 evaluated at two is equal to negative five and 𝑠 evaluated at five is equal to negative two. Another particle moves such that its position is given by ℎ of 𝑡 is equal to 𝑠 of 𝑡 plus 𝑡. Which of the following must be true? Option (A) ℎ has at least one zero. Option (B) 𝑠 has at least one zero. Option (C) 𝑠 of 𝑡 is greater than negative five and 𝑠 of 𝑡 is less than negative two for all values of 𝑡 between two and five. Option (D) There exists at least one 𝑐, where 𝑐 is greater than two and 𝑐 is less than five, such that 𝑠 evaluated at 𝑐 is equal to one. Or option (E) there exists at least one 𝑐, where 𝑐 is greater than two and 𝑐 is less than five, such that ℎ evaluated at 𝑐 is equal to five.

The question tells us the position of this particle moving along a line at time 𝑡 is given by the function 𝑠 of 𝑡 which is continuous. And we’re also told when 𝑡 is equal to two, our function 𝑠 of 𝑡 is equal to negative five. And when 𝑡 is equal to five, our function 𝑠 of 𝑡 is equal to negative two. This is all the information we’re given about our function 𝑠 of 𝑡. We’re also told there’s another particle whose position is given by the function ℎ of 𝑡, which is equal to 𝑠 of 𝑡 plus 𝑡.

We can use the fact that we know 𝑠 evaluated at two and 𝑠 evaluated at five to find ℎ evaluated at two and ℎ evaluated at five. We get that ℎ evaluated at two is negative three and ℎ evaluated at five is three. Also, since we know that our function 𝑠 of 𝑡 is continuous and 𝑡 is continuous, ℎ of 𝑡 which is 𝑠 of 𝑡 plus 𝑡 is the sum of two continuous functions. So, ℎ of 𝑡 must also be continuous.

So far, we’ve shown that both 𝑠 of 𝑡 and ℎ of 𝑡 are continuous functions. We were given two values for 𝑠 of 𝑡. And we’ll use this to find two values for ℎ of 𝑡. Let’s plot these four points onto a graph. We’re now ready to start going through our options. Option (A) tells us that ℎ must have at least one zero. The first thing we notice about our two points in the sketch of ℎ of 𝑡 is one lies below the 𝑥-axis and the other lies above the 𝑥-axis. And remember, we also know our function ℎ of 𝑡 is continuous. So, this should remind us of the intermediate value theorem.

We recall the intermediate value theorem tells us if a function 𝑓 is continuous on the closed interval from 𝑎 to 𝑏 and we have some value 𝑁 which is between 𝑓 evaluated at 𝑎 and 𝑓 evaluated at 𝑏. Then there exists a 𝑐 in the open interval from 𝑎 to 𝑏 such that 𝑓 evaluated at 𝑐 is equal to 𝑁. We want to apply the intermediate value theorem to our function ℎ of 𝑡 to show that it has at least one zero.

So, we’ll set our value of 𝑁 equal to zero, our function 𝑓 to be ℎ. And we’ll set the endpoints of our intervals to be the two points that we know for our function ℎ of 𝑡. That’s two and five. Now, our new version of the intermediate value theorem tells us if ℎ is continuous on the closed interval from two to five and zero is between ℎ evaluated at two and ℎ evaluated at five. Then there exists a 𝑐 on the open interval from two to five such that ℎ evaluated at 𝑐 is equal to zero.

In other words, if we can show that we’re allowed to use the intermediate value theorem for ℎ of 𝑡, then we can prove that there exists at least one zero. There are two conditions for using this version of the intermediate value theorem. First, we need to show that ℎ is continuous on the closed interval from two to five. And we already showed that this is true. ℎ of 𝑡 is the sum of two continuous functions, so it’s continuous.

The second condition we need to show is true is zero is between ℎ evaluated at two and ℎ evaluated at five. And of course, we already found these values; they’re negative three and three, respectively. And zero is between negative three and three. So, this condition is also true. So, both of the conditions for this version of the intermediate value theorem are true. This means there must exist a 𝑐 on the open interval from two to five such that ℎ evaluated at 𝑐 is equal to zero.

Therefore, option (A) must be true. ℎ must have at least one zero. We could stop there, but for due diligence, let’s check all of our other options. Option (B) tells us that our function 𝑠 must have at least one zero. To show that this is not true, we just need to find the continuous function which passes through both of these points and has no zeros. One such example could be the following piecewise linear function. We’ll define this function as negative two if 𝑡 is greater than or equal to five, 𝑡 minus seven if 𝑡 is between two and five, and negative five if 𝑡 is less than or equal to two.

We can see from this definition of 𝑠 of 𝑡 that 𝑠 of two is negative five and 𝑠 of five is negative two. It’s also a piecewise linear function where all of the endpoints meet up. So, 𝑠 of 𝑡 is continuous. And we can see that 𝑠 of 𝑡 has no zeros. Therefore, option (B) does not have to be true. So, now that we’ve shown that option (B) is not true, let’s clear some space and work on option (C).

Option (C) tells us we should be able to bound our function 𝑠 of 𝑡 between negative five and negative two for all values of 𝑡 between two and five. We can show that this is also false by using a similar method we did before. We’ll use a piecewise linear function. One such example of this is 𝑠 of 𝑡 is equal to negative 𝑡 over two minus four if 𝑡 is between zero and four. And 𝑠 of 𝑡 is equal to four 𝑡 minus 22 if 𝑡 is greater than four.

Again, we can see that 𝑠 evaluated at two is equal to negative five and 𝑠 evaluated at five is equal to negative two. However, we can also see that 𝑠 evaluated at four is equal to negative six. And 𝑠 of 𝑡 is a piecewise linear function whose endpoints meet. So, it’s a continuous function. So, we’ve again found a continuous version of 𝑠 of 𝑡 which passes through our two points. However, in our case, when 𝑡 is equal to four, 𝑠 of 𝑡 is negative six. So, it’s not between negative five and negative two when 𝑡 is between two and five.

We don’t actually need to clear our working for option (D). We can see that option (D) tells us that there should exist at least one 𝑐, where 𝑐 is between two and five, such that 𝑠 evaluated at 𝑐 is equal to one. However, if we look at our version of 𝑠 of 𝑡, where 𝑡 is between two and five, we can see the highest output of 𝑠 of 𝑡 is negative two. So, in this case, there’s no such 𝑐 such that 𝑠 of 𝑐 is equal to one. So, option (D) is also not true. Let’s clear our working and look at the final option, option (E).

Option (E) tells us that there should exist at least one 𝑐 between the values of two and five such that ℎ evaluated at 𝑐 is equal to five. One such way to show that this is not true is to take the straight line which passes between our two points, where 𝑡 is greater than or equal to zero. That’s ℎ of 𝑡 is equal to two 𝑡 minus seven. And remember that ℎ of 𝑡 is equal to 𝑠 of 𝑡 plus 𝑡. So, this also tells us that 𝑠 of 𝑡 is equal to 𝑡 minus seven. And of course, this version of 𝑠 of 𝑡 is linear. So, it’s continuous and it passes through our two points.

Let’s look at the largest output of ℎ of 𝑡, where 𝑡 is between two and five; it’s equal to three. If the highest output of our function is three in this interval, it can never be equal to five.

Therefore, we’ve shown for a continuous function 𝑠 of 𝑡, where 𝑠 of two is equal to negative five and 𝑠 of five is equal to negative two and the second function ℎ of 𝑡 which is equal to 𝑠 of 𝑡 plus 𝑡. Then, the only one of our options which must be true is that ℎ must have at least one zero.