# Question Video: Intermediate Value Theorem Mathematics • 12th Grade

Let 𝑓(𝑥) = 3^(𝑥) − 𝑥⁵. According to the intermediate value theorem, which of the following intervals must contain a solution to 𝑓(𝑥) = 0? [A] [2, 3] [B] [0, 1] [C] [−3, −2] [D] [1, 2] [E] [−2, −1]

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### Video Transcript

Let 𝑓 of 𝑥 equal three to the 𝑥 power minus 𝑥 to the fifth power. According to the intermediate value theorem, which of the following intervals must contain a solution to 𝑓 of 𝑥 equals zero? Is it the closed interval from two to three, the closed interval from zero to one, the closed interval from negative three to negative two, the closed interval from one to two, or the closed interval from negative two to negative one?

So we have a function. How can we use the intermediate value theorem to tell which of these intervals has a root of this function, a solution to 𝑓 of 𝑥 equals zero? Well, let’s remind ourselves of the intermediate value theorem. It states that if a function 𝑓 is continuous on the closed interval from 𝑎 to 𝑏 and 𝑁 is some number between the values of the function at the end points of that interval. That’s 𝑓 of 𝑎 and 𝑓 of 𝑏. Then there exists some number 𝑐 in the open interval from 𝑎 to 𝑏, such that 𝑓 of 𝑐 equals 𝑁. The first thing to notice that our function 𝑓 is continuous on the real numbers and so will be continuous on any of the intervals in the options as well. So this hypothesis holds.

Now remember, we want to find a solution to 𝑓 of 𝑥 equals zero. Comparing this with 𝑓 of 𝑐 equals 𝑁, it looks like we want to set 𝑁 equal to zero. So the intermediate value theorem is telling us that for our continuous function 𝑓, if zero is between 𝑓 of 𝑎 and 𝑓 of 𝑏, then there exists 𝑐 in the open interval from 𝑎 to 𝑏 such that 𝑓 of 𝑐 is zero. In other words, if 𝑓 of 𝑎 and 𝑓 of 𝑏 have different signs, then there is some number 𝑐 between 𝑎 and 𝑏 which is a root of 𝑓.

So to solve this question, we take each interval in the options in turn, starting with the interval from two to three. And if the signs of 𝑓 of two and 𝑓 of three are different, if one of them is positive and one of them is negative, then we know there must be a root, a solution to 𝑓 of 𝑥 equal zero in this interval. So let’s compute 𝑓 of two and 𝑓 of three. We do this using the definition for 𝑓 of 𝑥 we have in the question. We can evaluate these by hand or using a calculator, finding that 𝑓 of three is negative 216 and 𝑓 of two is negative 23. There’s no sign change of the function here. Both values are negative.

By the intermediate value theorem, we know that the function 𝑓 must take all values between negative the 23 and negative 216 as its input changes from two to three. So we’d have a solution to 𝑓 of 𝑥 equals negative 100. For example, in this interval. However, as zero is not between negative 23 and negative 216, we can’t say that there must be a solution to 𝑓 of 𝑥 equals zero in this interval.

We move on to option B. The closed interval from zero to one. We compute the values of the function at the end points. We find that 𝑓 of one is two and 𝑓 of zero is one. Again, there’s no sign change, so no guarantee of a zero in this interval. However, we can see a sign change between 𝑓 of one and 𝑓 of two. 𝑓 of one is positive and 𝑓 of two is negative. The intermediate value theorem tells us that as 𝑓 is continuous on the closed interval from one to two and as zero is between 𝑓 of one, which is two, and 𝑓 of two, which is negative 23. Then there exists a number 𝑐 in the open interval from one to two such that 𝑓 of 𝑐 is zero. And as 𝑐 lies in the open interval from one to two, it must also lie in the closed interval from one to two. And so we have a solution to 𝑓 of 𝑥 equals zero in the closed interval from one to two. This is option D.

We can — if we like — check the values of the function at the end point of the other intervals in the options to see that there is no sign change in either intervals C or E. And so, D is definitely the only correct answer. While the intermediate value theorem guarantees us a root or solution to 𝑓 of 𝑥 equals zero in the interval from one to two, we can’t say just based off the intermediate value theorem that there are no roots in the other intervals.