Video: Using the Intermediate Value Theorem to Find an Interval Containing a Solution to an Equation Involving Trigonometric Functions

According to the intermediate value theorem, in which of the following intervals must cos π‘₯ + sin π‘₯ + π‘₯ = 2 have a solution? [A] [1, 2] [B] [βˆ’2, βˆ’1] [C] [2, 3] [D] [βˆ’3, βˆ’2] [E] [0, 1]

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Video Transcript

According to the intermediate value theorem, in which of the following intervals must the equation cos π‘₯ plus sin π‘₯ plus π‘₯ equals two have a solution? Is it the answer A) the closed interval from one to two, B) the closed interval from negative two to negative one, C) the closed interval from two to three, D) the closed interval from negative three to negative two, or E) the closed interval from zero to one.

Let 𝑓 of π‘₯ equal cos π‘₯ plus sin π‘₯ plus π‘₯. Then, the question is asking us to use the intermediate value theorem to determine which of the listed intervals contains a solution to the equation 𝑓 of π‘₯ equals two. Let’s recall the intermediate value theorem. The intermediate value theorem says, suppose that 𝑓 is a continuous function on the closed interval from π‘Ž to 𝑏 and let 𝑁 be any number in the open interval from 𝑓 of π‘Ž to 𝑓 of 𝑏. Then, there exists a number 𝑐 in the open interval from π‘Ž to 𝑏 such that 𝑓 of 𝑐 equals 𝑁.

Let’s use the intermediate value theorem to check whether the closed interval from one to two, which is option A contains a solution to the equation 𝑓 of π‘₯ equals two. In order to do so, in the intermediate value theorem, let 𝑓 of π‘₯ equal cos π‘₯ plus sin π‘₯ plus π‘₯, π‘Ž equal one, 𝑏 equal two, and 𝑁 equal two.

The first step to using the intermediate value theorem to check whether the closed interval from one to two contains a solution to the equation 𝑓 of π‘₯ equals two is to check that 𝑓 is continuous on the closed interval from one to two. Since the functions 𝑦 equals cos π‘₯, 𝑦 equals sin π‘₯, and 𝑦 equals π‘₯ are all continuous on the set of all real numbers, so is their sum: 𝑓 of π‘₯ equals cos π‘₯ plus sin π‘₯ plus π‘₯. This means that, in particular, 𝑓 of π‘₯ is continuous on the subset of real numbers, the closed interval from one to two.

Next, we need to check if the number two lies in the open interval from 𝑓 of one to 𝑓 of two. If so, then the intermediate value theorem tells us that there exists a number 𝑐 in the open interval from one to two, such that 𝑓 of 𝑐 equals two. Note that if 𝑐 lies in the open interval from one to two, then 𝑐 also lies in the closed interval more from one to two. And so, we would have found a solution π‘₯ equals 𝑐 to the equation 𝑓 of π‘₯ equals two in the closed interval from one to two.

So, in order to check whether or not the closed interval from one to two is the correct answer to our question, all we need to do is check whether or not the number two lies in the open interval from 𝑓 of one to 𝑓 of two. Using a calculator, we find that 𝑓 of one is approximately 2.38 and 𝑓 of two is approximately 2.49. Remember to set your calculator to radian mode when doing these calculations. Since two does not lie between 2.38 and 2.49, two does not lie in the open interval from 𝑓 of one to 𝑓 of two.

And so, we cannot use the intermediate value theorem to deduce that the closed interval from one to two contains a solution to the equation 𝑓 of π‘₯ equals two. Note that this does not guarantee that the closed interval from one to two does not contain a solution to the equation 𝑓 of π‘₯ equals two. It just means that we cannot use the intermediate value theorem to guarantee a solution.

Let’s use the intermediate value theorem in a similar way to find which of the remaining intervals guarantees a solution to the equation 𝑓 of π‘₯ equals two. Firstly, remember that 𝑓 of π‘₯ is continuous on the set of all real numbers. And so, in particular, it is continuous on each of the remaining intervals. Now, all we need to do is evaluate the function 𝑓 at the 𝑁 points β€” π‘Ž and 𝑏 β€” of each of the remaining intervals and determine the open interval from 𝑓 of π‘Ž to 𝑓 of 𝑏 that contains the number two.

𝑓 of negative two is approximately equal to negative 3.33. 𝑓 of negative one is approximately equal to negative 1.3. Since two does not lie between negative 3.33 and negative 1.3, we deduce that we cannot use the intermediate value theorem to guarantee a solution to the equation 𝑓 of π‘₯ equals two in the closed interval from negative two to negative one. 𝑓 of two is approximately equal to 2.49 and 𝑓 of three is approximately equal to 2.15. Since two doesn’t lie between 2.49 and 2.15, we deduce that we cannot use the intermediate value theorem to guarantee a solution to the equation 𝑓 of π‘₯ equals two in the closed interval from two to three.

𝑓 of negative three is approximately equal to negative 4.13 and 𝑓 of negative two is approximately equal to negative 3.33. Since two does not lie between negative 4.13 and negative 3.33, we deduce that we cannot use the intermediate value theorem to guarantee a solution to the equation 𝑓 of π‘₯ equals two in the closed interval from negative three to negative two.

𝑓 of zero is equal to one and 𝑓 of one is approximately 2.38. Finally, we see that since two lies between one and 2.38, the intermediate value theorem guarantees the solution to the equation 𝑓 of π‘₯ equals two in the closed interval from zero to one. So, our final answer is that, according to the intermediate value theorem, the closed interval from zero to one contains a solution to the equation cos π‘₯ plus sin π‘₯ plus π‘₯ equals two. This corresponds to option E.

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