a. Use the Intermediate Value Theorem to show that the equatio...

Question

a. Use the Intermediate Value Theorem to show that the equation has a solution in the given interval.

x=cos x; (0,π/2)

Accepted Answer

Hello there. Today we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Determine if the equation sin of X minus X divided by 2 equals 0 has at least one solution in the interval, divided by 2. Using the intermediate value theorem. Awesome. So it appears for this particular problem, we're trying to determine whether or not this specific provided equation has at least one solution within this specific interval using the intermediate value theorem. Awesome. So now that we know what we're ultimately trying to solve for, let's read off our multiple choice answers to see what our final answer might be. A is the equation sin of x minus x divided by 2 equals 0 has at least one solution in the intervals divided by 2 according to the intermediate value theorem. The equation sin of x minus x divided by 2 equals 0 has no solution in the intervals divided by 2 according to the intermediate value theorem. Awesome. So, with that said, first off, we need to recall and note that the intermediate value theorem states that if F is a continuous function on the closed interval and let us know once again that a closed interval states that A. B is within 2 square brackets. So once again, the intermediate value theorem states that if F is a continuous function on the closed interval square bracket AB2 bracket. And N is any number between F of A and F of B. Then there exists at least one number C in the interval square bracket A B2, such that F of C equals N. So with that said, let us state that F of X is equal to sin of x minus X divided by 2. And we need to take this expression, and we need to check whether this given function expression is continuous on, so we're trying to figure out if it's continuous on parenthesis divided by 2, where it's important to note that these parentheses indicate that it's an open interval. So that's what a parenthesis means versus a square bracket where the square bracket indicates that it's a closed interval. So with that said, so we're now checking whether or not our given function is continuous on this specific interval of parenthesis pi divided by 2 parenthesis. To determine if the intermediate value theorem can be applied. So we need to recall and note that both sin of X and X divided by 2 will be continuous on this interval of pi divided by 2 pi within parentheses. Since sine is a periodic function that is continuous for all. All real numbers and its argument is a linear function that is defined for all real numbers. Therefore, as a result, the given function will be continuous on parenthesis pi divided by 2. Since the difference of the two continuous functions will also be continuous. And since we are asked to determine if the equation sine of X minus X divided by 2 equals 0 has at least one solution within the interval parenthesis divided by 2, we need to set N equal to 0, so let's take a note that we need to set N as equal to 0. Fantastic. And N equals 0 needs to lie between F of A. And F of B, so needs to lie between F of A and F of B. Fantastic. So, with that said, so since it has to lie between FFA and FAB, this will mean as a result that FFA and FAB must have opposite signs in order to prove the existence of a solution. So, now at this point, we need to evaluate the function at the endpoints of the interval and check if there is a sign change. So that means that we now need to evaluate F of X at the lower end point, so we're evaluating it at X equals pi divided by 2. Awesome. So now what we need to do, since we're evaluating at X equals pi divided by 2, we need to take F of X, which in this case, we need to take our X and we're plugging in a value of pi divided by 2, N for X. So you take F of pi divided by 2. is equal to sign of pi divided by 2 because once again, wherever there's an X value, we're plugging in pi divided by 2. And we need to subtract I divided by 2 divided by 2. And when we perform this like when we evaluate this expression, we will find that it will be equal to 1 minus pi divided by 4. Awesome. So now, our next step is we need to evaluate F of X, we're evaluating F of X at X equals pi. Fantastic. So now that we're evaluating at X equals pi, we need to take F of pi is equal to sign of pi. These we're plugging in a value of pi in for x minus pi divided by 2, which is gonna be equal to when we evaluate this expression, it's gonna be equal to simply negative pi divided by 2. Fantastic. So now, our next step is we need to check for a sign change between, so we're checking for a sign change between F of pi divided by 2, and F. Of pi. So we're trying to figure out once again if there's a sign change between F of pi divided by 2 and F of pi. So that will mean that at, so we're noting that at X equals pi divided by 2, we need to note that F of pi divided by 2 will be equal to 1 minus pi divided by 4, which is positive, and for X equals pi. We need to note that F of i will be equal to negative pi divided by 2, which will be a negative value. So now, at this point, we can now once again recall and apply the intermediate value theorem, and we need to note that since F of X changes sign from positive to negative, as X increases from pi divided by 2 to pi. According to the intermediate value theorem, there must be at least one solution, C in the interval parenthesis pi divided by 2 pi, such that F is C equals 0. So that means That without a doubt, our final answer has to be multiple choice answer letter A. And that's it. We've officially solved for this problem. Hooray, we did it. So once again, multiple choice answer letter A states that the equation sine of X minus X divided by 2 equals 0 has at least one. in the intervals divided by 2, according to the intermediate value theorem. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

a. Use the Intermediate Value Theorem to show that the equation has a solution in the given interval.x=cos x; (0,π/2)

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a. Use the Intermediate Value Theorem to show that the equation has a solution in the given interval.

x=cos x; (0,π/2)