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Let f(x) =x3 2x2 _ Sx + 6 be a fiunction defined on R Which of the following statement(s) is/are TRUE ?Newton' $ method is applicable to find the root of the c...

Question

Let f(x) =x3 2x2 _ Sx + 6 be a fiunction defined on R Which of the following statement(s) is/are TRUE ?Newton' $ method is applicable to find the root of the curve with the initial quess Xo It's possible to use Bisection method on [~1.5,0.5]. III: Fixed-point iteration can be applied to find the root of the 2x2+6 funetion with the initial guess Xo =1 for g(x)A.) Only IB.) Land IILand IIID.) Only IIIE.) I, I and III

Let f(x) =x3 2x2 _ Sx + 6 be a fiunction defined on R Which of the following statement(s) is/are TRUE ? Newton' $ method is applicable to find the root of the curve with the initial quess Xo It's possible to use Bisection method on [~1.5,0.5]. III: Fixed-point iteration can be applied to find the root of the 2x2+6 funetion with the initial guess Xo =1 for g(x) A.) Only I B.) Land II Land III D.) Only III E.) I, I and III



Answers

Determine whether the following statements are true and give an explanation or counterexample. a. Newton's method is an example of a numerical method for approximating the roots of a function. b. Newton's method gives better approximations to the roots of a quadratic equation than the quadratic formula. c. Newton's method always finds an approximate root of a function.

So we want to determine which of these statements is not true. Um, so if we divide this function by X squared plus one, we would get a remainder of seven as evidence by the plus seven at the end of the function. So this statement is true, Part B. If we have X minus two, we divide it, we will get seven in the same way we would divide by X squared, plus wanted get seven. So this statement is true for part C. It says the X minus f of to is equal to seven. Um, let's skip that one and go to D. So if we put D if we plug in zero, we would get one minus two, um, one times a negative, too. S That's negative. Two plus seven is five. That one's true. And then we look at, um, Part E. And it says that F does not have a really route and we see and that this would also be true, because if we were to graph this, um, it shows that there there is no real root. So this one's true. The one that we do wanna look at, though, is F of to being equal to seven? Um, actually, no. This one is here. Well, we'll come back to eat f of to being included. Seven. If we plug in a two for X value, we will end up getting zero plus seven. So that is accurate. This one's true. The one that we want to look at his part. E. I'm not saying the f does not have a real root, and that is not true, because all we would have to do is let f of X equal zero. Um, and we could solve for X, and we would end up getting values of X that, um, do, in fact, have riel routes. If you were to look at the graph of this function, you would see that it does intersect the X axis. That means that it does have a real route, which means that statement E is false.

So here we have a function which is a quality question which is a six X squared minus five x minus six. And there are some questions related to this expression and we need to answer that. Okay, so before starting it will just find the basic things like the discriminate of this equation, which will be 25 plus four and +26 and +26 It was just 169. So at 169 and Rudy comes out the victor. Okay, now moving on to the options first option says that the minimum value of fx is attained when X is equal to and X equals 25 by 12. Okay, it says this. Now for as minimum we know that the graph of function is something like this. Since these equals to zero. It will cut the graph, it will come to accept. So graph will be something like this. Also a is positive. Okay, six is positive. So graph will be upwards. So if you see this is the minimum point. This is called vortex. Okay. Vortex is the minimum point or the maximum points for the graph upwards or downwards respected. So for this graph, vortex is the minimum points. And we also know that coordinated vortex is given by minus b right away, coma minus D by. For These are the coordinates of vertex. Okay, so the minimum value will occur at -7 by the way. So minimum value for FX minimum. What we know that X will be minus B by two. Okay, so we know that um B is equal to five. Okay, B is given to B minus B is given to be five divided by 28 and 82 and 26 which is 55 Okay, so this is why by 12, which is the pro option. Okay, so the first option is through now moving on to the second option. Second option says that product of the roots. Second option asked about the product of the route alpha and beta which are given by sea by A And she is uh -6 and a six. So it obviously comes out to be -1. And then the question we are given it is equals 2 -1. So we can easily say that second option is also proof. Now let's see the third option part option says that it will not cut ex excess but we have just seen that the day of the graph is positive. Okay, the discriminate of this graph is positive. So it will obviously cut the ground. Obviously cut the excesses. We have just seen this graph. Okay, so corruption is false. Uh what does it mean that it cuts the excesses? That means it has real roots. Okay, whenever these greater than zero, the expression will have real roots. That means it will cut Xx is okay. And wherever it cuts excess excess excess, these are only the roots of the equation. Now let's move on to the fourth option which asked about domain of this function. Now. Not that there is nothing uh in the denominator of dysfunction. Okay, it is just simple expression which is 66 square minus five x minus six. There is nothing in the denominated. So no, no value of X. Can make it and determinate. Okay, no value of X can bring it into the indeterminate form. So domain of fX is uh the X. Belongs to our. So this is also too. So these are the answers of the older four parts. Thank you.

The only statement here that is both is e these rules. The reason being is that the number of non real complex heroes of us could be one that's completely false because you need even number non riel, complex liberals. You could have 02 works on 68 but not just mutton.

Given 1/3 degree polynomial off of X equals April's be exposed six scored post e x Cubed D does not equal zero. Then we know that after double prime of X equal to zero, we would have given the same variables X equals negative to see over sex d which is equivalent to negative. See over 30. As you can see with the side of second derivative changes at X equals Night of Sea over three D. Therefore, there is always going to be an inflection point because when the scientist I contributed changes, we have an inflection point. So the answer to this is gonna be false. Inflection point would be at sequels to over three times. They're this cancels out in his one to over one equals two.


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