Encuentra la raíz de la función h(x) = x^4

Encuentra la raíz de la función h(x) = x^4 - 2x^3 + x^2 - 3 en el intervalo [-2, -1], utilizando el método de bisección. Encuentra la raíz de la función g(x) = cos(x) - x en el intervalo [0, 1], utilizando el método de bisección.

Transcript

00:01 Hello and welcome to this video solution of numerate.

00:04 So here, question 1, you have to evaluate the root of this function between this interval minus 2 to minus 1 using the method of bi -section.

00:13 So first of all, what you need to calculate is f of h of minus 2, which is minus 2 to the power of 4, minus 2 minus 2 to the power of cube 3, plus this is minus 2 square minus 3, right? and that gives you 33 and next is h of minus 1 you have to calculate which is minus 1 to the bar of 4 minus 2 times of minus 1 cube plus this is minus 1 square plus minus 3 right and that is equal to 1 itself right so what we see is that these two values h of minus 2 and h of minus 1 have same sign right both are positive if you see both are plus right so both are having same signs so no root exist between them so next question two now here what you can do is in the next question g x is cos x minus x and the interval is 0 to 1 right so first of all you have to start with g 0 that is equal to 1 right and g 1 you have to take that is equal to minus 0 .4597 right so next if you see here the signs are opposite so this is plus and here you have got minus so you have to find out see the midpoint between them c1 let's say first iteration first iteration c1 that is equal to 0 plus 1 by 2 and that is 0 .5 now you have to calculate g of 0 .5 and that's here you have got cost of 0 .5 minus of 0 .5 right and that is equal to 0 .376 and now this is positive right so this is positive and one is negative so here second iteration you have to take which is c2 that is equal to 1 plus 0 .5 by 2 so it's 0 .75 so the root lies between both of them now you have to calculate g of 0 .6 you can do it like minus 0 .0183 right so this is negative and you have got 1 as positive here sorry 0 .5 as positive and this is negative so third iteration you have to take c3 to be equal to 0 .5 and 0 .75 because the root will lie between them by 2 and this is 0 .625 right now here you can take g of 0 .625 that is equal to 0 .1859 right 4th iteration you have to take c4 now this is positive and you have to see okay 0 .75 is negative right so c4 is 0 .75 plus 0 .625 by 2 and that is around 0 .6875 right and you have to take g of 0 .6875 again that is equal to 0 .0855 right so this is positive and here you've got positive and 0 .75 is negative so fifth iteration you have to do a lot of iterations here actually c5 that is equal to 0 .7 plus 0 .6875 by 2 that is equal to 0 .71875 right okay so you can proceed it a bit further and you can continue like that and g of 0 .71875 equal to something and you can proceed a fit further or you can stop here itself right which you can feel right you can proceed on maybe we'll just solve it the final one and this value is 0 .0 .03355 right but yeah this is positive and next is now where is the negative one let me see 0 .75 is negative so yeah the same c6 6 6 iteration c6 you can just take like 0 .71875 plus 0 .75 by 2 and that's around 0 .7442 and 73 .7 .731875...

Question

Encuentra la raíz de la función h(x) = x^4 - 2x^3 + x^2 - 3 en el intervalo [-2, -1], utilizando el método de bisección. Encuentra la raíz de la función g(x) = cos(x) - x en el intervalo [0, 1], utilizando el método de bisección.

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