{Use of Tech} Finding all roots Use Newton’s method to find al...

Question

{Use of Tech} Finding all roots Use Newton’s method to find all the roots of the following functions. Use preliminary analysis and graphing to determine good initial approximations.

f(x) = cos 2x - x² + 2x

Accepted Answer

Below 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. Use Newton's method to find all roots of F of X is equal to cosine of X minus X2 + X by first using analysis or graphing to choose good starting values. Hint one root lies between -1 and 0, and another between 1 and 2. OK, so it appears for this problem we're asked to use Newton's method to help us find all the roots of our function F of X, and we're asked to do this by first using analysis or a graphing method to help us determine what our good starting values should be to help us determine or choose good starting values. We're also Given a hint that tells us that one root lies between the range -1 and 0 and another one between 1 and 2. So with that said, first off, let us quickly note once again that our function that we are focusing our attention on is going to be F of X is equal to cosine of X minus X2 + X. And when we take this function and we plug it into our graphing calculator or whatever graphing software makes sense for us. When we use, or whatever graphing software makes the most sense for us to use, we should get the following graph when we go ahead and just take our graph of FFX and we plug it directly into our graphing software. We should receive the following graph, which appears to be parabolic in nature. We have a parabolic curve, which appears to be an inverted parabola, which is represented in blue. Our curve is in baby blue, and it's on our XY coordinate plane. Fantastic. And as we can see from our graph of F of X, we can therefore denote that it is true that one root lies between the range of -1 and 0, and another root lies between the range of 1 and 2, as we can see from our graph. So therefore, we can easily recall and use the iterative formula for Newton's method, which as we should try, like since we need to use, so ultimately we're gonna have to do iterations, and we need to recall and use the formula that we need to use to help us produce or solve for those iterations. So, as you should recall, Newton's method or the equation for Newton's method states that X subscript N + 1. is equal to x N minus F of X N divided by F of X N, or let us recall a note that F of X denotes the first derivative of our function F of X. So that's what our prime symbol means. The prime symbol means we're taking the first derivative of F of X. So, with that said, We now need to take the derivative of F of X. So when we take the derivative of F of X, we will find that F of X is going to be equal to negative of X minus 2 X + 1. Fantastic. So once again, that's when we take the first derivative of our function F of X, which once again F of X is equal to cosine of X minus X2 + X, and the first derivative of F of X is going to be equal to negative sin of x minus 2 X + 1. So therefore, as a result, our iterative function or iteration will become X N + 1 is equal to X N minus cosine of X N minus X N to the power of 2 plus X N divided by negative of X N minus 2 multiplied by X N + 1. Fantastic. So now we need to focus our attention now for the root that is in So now we're trying to focus our attention on the root that's in parentheses -10. So note that these parentheses means that it's an open range. If they were square brackets, then that means they'd be a closed range. So we're trying to find our root. Which is parentheses -1.0, and we're also told that initially, our initial guess X0 is equal to negative. 0.50. So with that said, we can now solve for our first iteration. So X1 is going to be equal to, so now for our value for X and so we're using our iteration formula, and we're going to be plugging in a value of negative 0.50 in for X subscript N. So when we do that, we will find that X1 is equal to -0.50. Minus cosine of -0.50. Minus -0.50 to the power of 2. Plus 0.50. All divided by negative sign of. So we have to take negative sign of -0.50, and then we need to subtract 2 multiplied by -0.50. And then we don't forget to add one more. So when we plug in this big old long expression into our calculator, making sure we have our parentheses in its proper locations, we will find when we round to 5 decimal places, we will receive an answer of negative 0. 55146. And once again, that's when you plug in this big old long expression into your calculator and you round to five decimal places, you'll get negative 0.55146 as your answer. So I don't have to write our iteration formula over and over again. We're just going to plug in our different values of X subscript N. And for our different iteration reason, I'm just gonna give you the final answer for the iteration. So now, what we need to do to solve for X subscript 2, our second iteration, we need to take a value of negative 0. 55146 in for XN and when we do that, we will find that we will produce an answer of negative 0.55001 as an answer. Hooray, we did it, and once again, that's when you plug in a value of negative 0.55146 in for X subscript N. So now we need to iterate again because we're trying to find two approximations that are equal to one another. So we need to plug in for our third iteration we need to plug in a value of -0.55001. And for our value of X subscript N, and when we do that, we will find when we round to 5 decimal places, we will get an answer of negative 0.55001. So that means, without a doubt, the two successive approximations within 5 decimal places when being rounded up that are equal to each other, were from the 2nd and 3rd iterations. So that means without a doubt, our route 1 is going to be equal to -0.55001, and that is our first answer. Hooray, we did it. Now we need to solve for our second root. So once again, this is our root that is in parentheses 12, fantastic. And now we need to note that we're gonna say our initial guess, let's say that X0 is equal to 1.50. Fantastic. So, similarly, like we did above, let's solve for X1, so our first iteration, which is gonna be equal to 1.50 because we're plugging in that value of 1.50 in for X subscript N, so it's going to be minus cosine of 1.50. So 1.50 minus 1.50 to the power 2, plus 1.50, fantastic. And then we need to divide everything by negative sign of 1.50. Minus 2 multiplied by 1.50 plus 1. So when we plug in that big old long expression into our calculator, we will find when we round to 5 decimal places that it's going to be equal to 1.27339. Fantastic. So now, like we did before, now we need to solve for our second iteration, so plugging in a value of 1.27339 in for X subscript N will give us an answer of 1.25138. As an answer, and so we don't get confused, let's move it down a little bit because that's our second iteration answer. Now we need to solve for our third iteration, and our third iteration. Once again, we need to plug in a value of 1.25138 in for X subscript N. And when we do that, we will get an answer of 1.25115 when we round to 5 decimal places. And so far we haven't had two successive approximations equal each other, so we need to do 1/4 iteration. So X subscript 4, so X4 is going to be equal to. So now we need to plug in a value of 1.25115 in for X subscript N. And when we do that, we will find that our final answer will be equal to 1.25115, when we round to 5 decimal places. So that means we have finally successfully found two successive approximations that are equal to one another. So therefore, the two successive approximations with 5 decimal places rounded up being equal to each other, were from the 3rd and 4th iterations. So that means that our route 2 is going to be equal to 1.25115, and that is our second answer. Hooray, we did it. So going back to the top, our final two answers without a doubt, will have to be Drum roll, 0.55001 and Positive 1.25115, and that's it. Those are our final two answers. Hooray, we did it. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

{Use of Tech} Finding all roots Use Newton’s method to find all the roots of the following functions. Use preliminary analysis and graphing to determine good initial approximations.f(x) = cos 2x - x² + 2x

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{Use of Tech} Finding all roots Use Newton’s method to find all the roots of the following functions. Use preliminary analysis and graphing to determine good initial approximations.

f(x) = cos 2x - x² + 2x