[Technology Exercise] Roots

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Question

[Technology Exercise] Roots

Let ƒ(𝓍) = 𝓍³ ―𝓍― 1.

b. Solve the equation ƒ(𝓍) = 0 graphically with an error of magnitude at most 10⁻⁸ .

Accepted Answer

Welcome back, everyone. In this problem, consider the graph of the function F X equals 3X cubed minus 2X minus 2. Approximate the root of the given function using Newton's method. A says it's 1.222706. B, 1.122706, C 1.123285, and D, 1.223285. Now, how can we approximate the route using Newton's method? What do we know about Newton's method? Well, first of all, recall. The Newton's method, OK. Newton's method. Basically tells us that XN + 1, the next approximation, is equal to xn, the current approximation minus F of XN, that is the function whose root is being approximated, divided by the derivative of F of XN. Where the derivative, OK. Is not equal to 0. Now, what that means then is that we'll need to find an initial guess XN OK, which we can call X0. And from that initial guess for our root, we can continue to repeat the process for a new turns method until the XN values converge to a root. So first of all, let's try to find a value for XN and let's also find the derivative of F of XN so that we can plug XN into that derivative. No, we can use our graph to help us get our initial guess. Notice here that based on our given graph, the root, OK, is somewhere between 1 and 1.25 on our X-axis, OK? So we can let our initial value X not be equal to 1.1. Also, we can find the derivative of FF X, OK? Where we know that FF X is 3X cubed minus 2X minus 2. So it's derivative will be 9 X2 minus 2. Now, we can go ahead and find our first approximation X1. Cause no X1 will be equal to X0 minus F of X0. Divided by the derivative of a perfect knot. So, now, in this case, that's gonna be equal to 1.1 minus F of 1.1. That's 3 multiplied by 1.1 cubed, minus 2 multiplied by 1.1 minus 2, OK. All divided by the derivative of F of 1.1, so that's 9 multiplied by 1.1 squared minus 2. I know when we go ahead and compute, we should get X1 to be approximately equal to 1.123285. Let's repeat the process for X2. No, we should get X2 then to be equal to X1 minus F of X1. Divided by the derivative of F of X1. And now in this case, we know that X1 is approximately 1.123285. So if we substitute that here, then we should get X2 to be approximately equal to 1.122707. Next, X3 will be equal to X2 minus F of X2 divided by the derivative of FF X2, and since it's approximately 1.122707, then X3 is going to be approximately 1.122706. Next, let's find X4. So now, that's gonna be X3 minus F of X3 divided by the derivative of F of X3. And by now you realize that we're just repeating the process as we had before. We're using, we're just substituting different values of X into our expression or, or our equation rather. And no for X4, then you should also find that it equals 1.222706. What do you notice? Well, here, notice that the 3rd and 4th iterations show that the values of XN converge to 1.122706. Therefore, the root is approximately equal to 1.122706. If we look back at our answer choices, that means B is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

[Technology Exercise] Roots

Let ƒ(𝓍) = 𝓍³ ―𝓍― 1.

b. Solve the equation ƒ(𝓍) = 0 graphically with an error of magnitude at most 10⁻⁸ .

Key Concepts

Roots of a Function

Graphical Method for Finding Roots

Error Tolerance in Numerical Methods

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