# Question Video: Using the Intermediate Value Theorem to Determine the Properties of a Function Mathematics • Higher Education

Let 𝑓(𝑥) = 𝑥 + sin 𝑥. According to the intermediate value theorem, which of the following must be true? [A] There exists at least one 𝑐, where −𝜋/2 < 𝑐 < 0, such that 𝑓(𝑐) = 1. [B] None of the choices is correct. [C] There exists at least one 𝑐, where 𝜋/2 < 𝑐 < 𝜋, such that 𝑓(𝑐) = 1. [D] There exists at least one 𝑐, where 0 < 𝑐 < 𝜋/2, such that 𝑓(𝑐) = 1. [E] There exists at least one 𝑐, where −𝜋 < 𝑐 < −𝜋/2, such that 𝑓(𝑐) = 1.

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### Video Transcript

Let 𝑓 of 𝑥 be equal to 𝑥 plus the sin of 𝑥. According to the intermediate value theorem, which of the following must be true? Option (A) there exists at least one 𝑐, where 𝑐 is greater than negative 𝜋 by two and 𝑐 is less than zero, such that 𝑓 evaluated at 𝑐 is equal to one. Option (B) none of the choices is correct. Option (C) there exists at least one 𝑐, where 𝑐 is greater than 𝜋 by two and 𝑐 is less than 𝜋, such that 𝑓 evaluated at 𝑐 is equal to one. Or option (D) there exists at least one 𝑐, where 𝑐 is greater than zero and 𝑐 is less than 𝜋 by two, such that 𝑓 of 𝑐 is equal to one. Or option (E) there exists at least one 𝑐, where 𝑐 is greater than negative 𝜋 and 𝑐 is less than negative 𝜋 by two, such that 𝑓 evaluated at 𝑐 is equal to one.

The question gives us a function 𝑓 of 𝑥 and four possible uses of the intermediate value theorem on our function 𝑓 of 𝑥. We need to determine which of these statements, if any, must be true by using the intermediate value theorem. So let’s start by recalling what the intermediate value theorem tells us. The intermediate value theorem says if 𝑓 of 𝑥 is a continuous function on the closed interval from 𝑎 to 𝑏 and the value of 𝑁 is between 𝑓 evaluated at 𝑎 and 𝑓 evaluated at 𝑏. Then the intermediate value theorem tells us there must exist a 𝑐 on the open interval from 𝑎 to 𝑏, where 𝑓 evaluated at 𝑐 is equal to 𝑁.

Let’s compare this statement to the answers given to us in the question. First, the concluding statement of the intermediate value theorem tells us about a value 𝑐 where 𝑓 of 𝑐 is equal to 𝑁. We can see that options (A), (C), (D), and (E) all tell us about the existence of a 𝑐 where 𝑓 of 𝑐 is equal to one. So our value of 𝑁 in the intermediate value theorem will be one. Let’s now check what we need to use the intermediate value theorem. We need our value of 𝑁 to be between the value of 𝑓 of 𝑎 and 𝑓 of 𝑏. However, if we look at each of our options, we have different values for 𝑎 and 𝑏 for all of our options.

It’s also worth pointing out we technically need to check 𝑓 of 𝑥 is continuous on each of these closed intervals. However, in our case, 𝑓 of 𝑥 is 𝑥 plus the sin of 𝑥. This is a linear function plus a trigonometric function. So in this case, 𝑓 of 𝑥 is continuous for all real numbers. So we don’t need to check this for each of our options. Now, to check the intermediate value theorem, we need to evaluate 𝑓 for each of our possible pairs of 𝑎 and 𝑏. We’ll start by finding 𝑓 evaluated at negative 𝜋 by two.

So if we substitute 𝑥 is equal to negative 𝜋 by two into our expression for 𝑓 of 𝑥, we get negative 𝜋 by two plus the sin of negative 𝜋 by two. And if we evaluate this expression, we get approximately negative 2.57. However, we’re only interested in “is this bigger or smaller than one?” This is less than one. We’ll now do the same for 𝑓 evaluated at zero. Substituting 𝑥 is equal to zero, we get zero plus the sin of zero, which we can evaluate is equal to zero. And this is also less than one. So in the case of option (A), our value of 𝑁 does not lie between 𝑓 evaluated at negative 𝜋 by two and 𝑓 evaluated at zero, since both of these are less than one. Therefore, option (A) cannot be correct.

Let’s now do the same for 𝜋 by two. Substituting this into our expression for 𝑓 of 𝑥, we get 𝜋 by two plus the sin of 𝜋 by two. And if we calculate this, we get approximately 2.57, which we know is bigger than one. And now we see 𝑓 evaluated at zero is less than one, but 𝑓 evaluated at 𝜋 by two is greater than one. So our value of 𝑁 lies between 𝑓 evaluated at zero and 𝑓 evaluated at 𝜋 by two. This means we can use the intermediate value theorem with 𝑓 of 𝑥 equal to 𝑥 plus the sin of 𝑥, 𝑎 equal to zero, 𝑏 equal to 𝜋 by two, and 𝑁 equal to one.

We see that 𝑓 of 𝑥 is continuous on the closed interval from zero to 𝜋 by two. We see that 𝑁 is between 𝑓 evaluated at zero and 𝑓 evaluated at 𝜋 by two. This then tells us there exists 𝑐 in the open interval from zero to 𝜋 by two where 𝑓 of 𝑐 is equal to one. Of course, 𝑐 being in the open interval from zero to 𝜋 by two means that 𝑐 is greater than zero and 𝑐 is less than 𝜋 by two. And we can see this is exactly what’s said in option (D). Therefore, we can in fact conclude option (D) by using the intermediate value theorem.

Now we could stop here, but let’s also check options (B), (C), and (E). Well, we’ve already shown that one of our options is correct. So option (B) can’t be correct. To check option (C), we want to find 𝑓 evaluated at 𝜋. This is equal to 𝜋 plus the sin of 𝜋. And the sin of 𝜋 is zero, so this is just equal to 𝜋. And this is greater than one. And now we see that 𝑓 evaluated at 𝜋 by two and 𝑓 evaluated at 𝜋 are both greater than one. So our value of 𝑁 can’t be between both of these. So option (C) can’t be concluded by using the intermediate value theorem.

And finally, to check option (E), we need to find 𝑓 evaluated at negative 𝜋. This is equal to negative 𝜋 plus the sin of negative 𝜋, which we can evaluate is equal to negative 𝜋. And of course, this is less than one. So again, both 𝑓 evaluated at negative 𝜋 and 𝑓 evaluated at negative 𝜋 by two are less than one. So one can’t be between these two values. And so option (E) also can’t be deduced by using the intermediate value theorem.

Therefore, if we’re told that a function 𝑓 of 𝑥 is equal to 𝑥 plus the sin of 𝑥. We can show by using the intermediate value theorem there must exist at least one 𝑐 where 𝑐 is greater than zero and 𝑐 is less than 𝜋 by two, such that 𝑓 evaluated at 𝑐 is equal to one.

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