maths help please!!! (1 Viewer)

boredsatan · Apr 30, 2017

The graph of y = x^4 - 2x - 12 has 2 x-intercepts
a. construct a table of values for this polynomial rule for x = -3,-2,-1,0,1,2,3
b. Hence state an exact solution to the equation x^4 - 2x - 12 = 0
c. State an interval within which the other root of the equation lies and use the methods of bisection to obtain an estimate of this root correct to 1 decimal place

I get how to do part a and b, but i'm finding part c extremely challenging and confusing.

thanks!!!

InteGrand · Apr 30, 2017

boredsatan said:
The graph of y = x^4 - 2x - 12 has 2 x-intercepts
a. construct a table of values for this polynomial rule for x = -3,-2,-1,0,1,2,3
b. Hence state an exact solution to the equation x^4 - 2x - 12 = 0
c. State an interval within which the other root of the equation lies and use the methods of bisection to obtain an estimate of this root correct to 1 decimal place

I get how to do part a and b, but i'm finding part c extremely challenging and confusing.

thanks!!!

If you know a place where the polynomial is positive and another place where it is negative (and the root you already found does not lie between these two places), then the other root lies between these two numbers. You can then use the bisection method (check your textbook if you haven't learnt it yet).

boredsatan · Apr 30, 2017

Problem is i'm still not understanding (for whatever reason it's just not making sense to me) the bisection methods even after reading my textbook and the summary you linked me

Mahan1 · Apr 30, 2017

boredsatan said:
Problem is i'm still not understanding (for whatever reason it's just not making sense to me) the bisection methods even after reading my textbook and the summary you linked me

$\textrm{In order to understand the bisection method you need to understand the following idea:}$

$\textrm{Let's use an easy example} \ y=x^{2}-9 \ \textrm{on the interval [1,7]}$

$\textrm{at} \ x=1, y=-8< 0 \ \textrm{but at} \ x=7, y=40 > 0 \ \textrm{so we have two points, A=(1,-8) and B=(7,40)}$

$\textrm{That means A is under x-axis and B is located above the x-axis. Since these points are connected.}$

$\textrm{If you try to connect A to B, you have to cross the x-axis somewhere on the x-axis.}$

$\textrm{So you at least know the function has a root somewhere between 1 and 7}$

$\textrm{The bisecting method is using this idea over and over again}$

$\textrm{Let's consider the middle of the interval [1,7] }$

$\textrm{which is at } \ x=\frac{1+7}{2}=4. \ \textrm{At} \ x=4,y=7 > 0$

$\textrm{Let's call that point corresponding to the middle point is}, C=(4,7). \ \textrm{Now You have three point that lie on the graph}$

$\textrm{Since A is below and C is above the x-axis, using the same argument, we have to conclude:}$

$\textrm{There is at least one root between A and C, since A=(1,-8),C=(4,7)}$

$\textrm{then the root of the function lies somewhere between 1 and 4 }$

$\textrm{Then you can use the bisecting method on the interval [1,4]}$

$\textrm{which you proved there is at least a root in that interval, again}$

$\textrm{you might find it useful to draw the approximate location of each point to keep track of the points}$

boredsatan · Apr 30, 2017

Mahan1 said:
$\textrm{In order to understand the bisection method you need to understand the following idea:}$

$\textrm{Let's use an easy example} \ y=x^{2}-9 \ \textrm{on the interval [1,7]}$

$\textrm{at} \ x=1, y=-8< 0 \ \textrm{but at} \ x=7, y=40 > 0 \ \textrm{so we have two points, A=(1,-8) and B=(7,40)}$

$\textrm{That means A is under x-axis and B is located above the x-axis. Since these points are connected.}$

$\textrm{If you try to connect A to B, you have to cross the x-axis somewhere on the x-axis.}$

$\textrm{So you at least know the function has a root somewhere between 1 and 7}$

$\textrm{The bisecting method is using this idea over and over again}$

$\textrm{Let's consider the middle of the interval [1,7] }$

$\textrm{which is at } \ x=\frac{1+7}{2}=4. \ \textrm{At} \ x=4,y=7 > 0$

$\textrm{Let's call that point corresponding to the middle point is}, C=(4,7). \ \textrm{Now You have three point that lie on the graph}$

$\textrm{Since A is below and C is above the x-axis, using the same argument, we have to conclude:}$

$\textrm{There is at least one root between A and C, since A=(1,-8),C=(4,7)}$

$\textrm{then the root of the function lies somewhere between 1 and 4 }$

$\textrm{Then you can use the bisecting method on the interval [1,4]}$

$\textrm{which you proved there is at least a root in that interval, again}$

$\textrm{you might find it useful to draw the approximate location of each point to keep track of the points}$

There's nothing in your post

InteGrand · Apr 30, 2017

boredsatan said:
There's nothing in your post

I think he wrote it in LaTeX, but it's just been lagging on this site recently.

boredsatan · Apr 30, 2017

Will it be working after a few hours?

InteGrand · Apr 30, 2017

boredsatan said:
Will it be working after a few hours?

Not sure. I think it's been lagging like this for the past few days.

maths help please!!! (1 Viewer)

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