Let ƒ(x) = 3x - x³ . Show that the equation ƒ(𝓍) = -4 ha...

Question

Let ƒ(x) = 3x - x³ . Show that the equation ƒ(𝓍) = -4 has a solution in the interval [2,3] and use Newton’s method to find it.

Accepted Answer

Hello there. Today we're going to 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. Consider the function G of X is equal to 2 x 2 minus 5 X + 1. By the intermediate value theorem, it can be shown that the equation GFX equals 2. As a solution in the interval, square bracket 2.3 square bracket, use Newton's method to find the solution in this interval. Approximate your answer to 3 decimal places. Fantastic. So it appears for this particular problem, we're asked to use Newton's method to help us find the solution within this specific closed interval, because that's what these square brackets mean. It's a closed interval of 2.3. And we're asked to use Newton's method to help us find the solution within this interval, and we're asked to approximate our answer to 3 decimal places. Fantastic. 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, noting that they all state that X is approximately equal to some value. So A is 2.632, B is 2.686, C is 2.788, and D is 2.837. OK. So first off, let us recall and use the intermediate value theorem. So as we should recall, the intermediate value theorem states that if a function F of X is continuous on a closed interval of A B in square brackets, then that will mean that And or we need to note that. Once again, so we're trying to note, we need to recall note that the inter intermediate value theorem states that if a function F of X is continuous, so I'll put a letter C for continuous. If it's continuous on a close interval of square brackets AB and that N N to be precise is any number between F of A. And F of B, then there exists at least one number, lowercase c in the interval parenthesis A B, so it's an open interval, A B, such that F of C is equal to N. So let's apply this theorem to our function. G of X is equal to 2. X to the power 2 minus 5 X plus 1, fantastic. And let us note that it's on the closed interval of square brackets +23. To show that it takes the value 2 somewhere on this interval. So we need to show that this takes the value of 2 somewhere within this interval. So now we need to evaluate G of X at the end points of the interval. So we need to calculate G of 2. So G. 2 is going to be equal to 2 multiplied by 2 to the power of 2 minus 5 multiplied by 2, plus 1, which is going to be equal to -1 when we plug in that expression in our calculator. Once again, we're plugging in a value of 2 in for X. So now we need to calculate G of 3. So plugging in a value of 3 in for X, we will get 2 multiplied by 3 to the power of 2, minus 5 multiplied by 3, plus 1, which will give us an answer of 4. So now, we need to note that we will have that GF2 is equal to -1, and we will find that GF3 is equal to 4. So since GFX is a quadratic polynomial, that means it will be continuous on the entire real line, including our interval of square brackets 3, 2.3. So this 2.3 is this closed interval of 2.3. That means that it's going to be continuous on the entire real line within this interval. So the value 2 between -1 and 4, so therefore by the intermediate value theorem, there must be some value is C. Which is an element of 2.3, such that G of C is equal to 2. Fantastic. So once again, C is an element of the open interval 2.3 and GFC is equal to 2. So for the function, G of X is equal to 2 x is the power 2 minus 5 X plus 1. We want to find a root of the equation. G of X is equal to 2. So that means we need to solve for H of X is equal to 2 X is the power 2 minus 5 X plus 1. Minus 2 is equal to 0. So now we need to simplify H of X. So H of X is going to be equal to 2x of the power 2 minus 5 X plus 1 minus 2. So when we simplify, we will find that H of X is equal to 2 X of the power 2 minus 5 X plus 1 minus 2, which is all equal to 2 X of the power 2 minus 5 X minus 1. Fantastic. So now we need to use the iterative. Formula for Newton's method. So we need to find the iteration formula for Newton's method. So using the inter interva since we need to take the interval. Innervation. So we're trying to find the innervation formula for Newton's methods. So as we should recall, that states that X to the subscript of, so X N + 1 is equal to X N minus H of X N divided by H X N. So now we need to find the first derivative of H of X. So H of X is equal to 2 X to the power of 2, so we need to note that H of X is equal to 2 X to the power of 2 minus 5 X minus 1. So when we take the first derivative of H of X, we will find that H of X is equal to 4 X minus 5. So that will mean That we can now apply Newton's method, but we need to use an initial guess X0. So a reasonable choice will be the midpoint of the interval square brackets 2.3, which in this case our educated guess that's within this close interval will be X0 as equal to 2.5. So let us perform a few iterations to help us approximate our route. So for X1. We need to note that X0 is going to be equal to 2.5, so we can say that X1 is equal to X0 minus H of X0 divided by H X0. So that will mean when we plug in our known variables, we will get when we plug in a value of 2.5 in for x0, we'll get 2.5. Minus 2 multiplied by 2.5 square minus 5 multiplied by 2.5. -1, all divided by 4 multiplied by 2.5. -5, so that will mean that X1, when we plug in that expression into our calculator and we round to one decile place, it will give us 2.7. So now we need to perform our second iteration. So our second iteration, so X1, we need to note is equal to 2.7. So now we need to solve for X2 X subscript 2. And let us note that X subscript 2 is going to be equal to X 1 minus H of X1 divided by H of X1. So when we plug in our known variable which is 2.7 in for X1, and I'm gonna not write the equation out again because It'll also save some time, so just note we're plugging in a value of 2.7 in for X1. So when we do that, we will find that X2 is approximately equal to 2.68620689, so it's gonna be 06897. So now we need to perform a 3rd iteration, because so far all the iterations that we've done, we haven't found two approximations that are equivalently equal to each other. So with that said, we need to perform a 3rd iteration. So when we do our 3rd iteration, we will find that X subscript 3 is going to be equal to X, subscript 2 minus H of X subscript 2 divided by HX subscript 2. Fantastic. So with that said, when we plug in our known interval, which is we're going to be plugging in a value of 2.686206897 in for X2, and when we do that, we will find that it's approximately equal to 2.686140663. So that will mean that X subscript 3 is approximately equal to me round to three decimal places, 2.686. So that will mean that without a doubt, since the approximated values in the 2nd and 3rd iteration agree to three decimal places, we can stop at this point and we can take 2.686 as our approximate solution, which is accurate to 3 decimal places. So that is our final answer. Hooray, we did it. So going back to look at our multiple choice answers. The correct answer, without a doubt has to be multiple choice answer letter B, which once again is X is approximately equal to 2.686. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Let ƒ(x) = 3x - x³ . Show that the equation ƒ(𝓍) = -4 has a solution in the interval [2,3] and use Newton’s method to find it.

Key Concepts

Intermediate Value Theorem

Newton's Method

Derivatives

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