# Using the Bisection Method

• Mar 3rd 2010, 07:15 PM
frenchguy87
Using the Bisection Method
let $f(x)=(x-1)(x-2)(x-3)(x-4)(x-5)$. This function has five roots on the interval $[0,7]$. I have to use the bisection method on this interval and figure out what root is located.
I'm not entirely sure I know how to use this method correctly..
Here it goes,
First, since $f(0)<0$ and $f(7)>0$, there exists a $c\in\mathbb{R}$ in between those two points such that $f(c) = 0$ ..
Now would you divide the interval in two?
• Mar 3rd 2010, 07:48 PM
pickslides
This is quite a strange function to use with this method as it is fully factored and gives you the roots.

Anyhow you choose the interval $[0,7]$ and find the signs of $f(0) <0$ and $f(7)>0$
Now bisect the interval $\frac{7-0}{2} = 3.5$ to make 2 smaller intervals and find the sign of $f(3.5)$

We know $f(3.5)>0$ so we choose the smaller intervals with opposite signs $f(0) <0$ and $f(3.5)>0$ , bisect again and so on until the interval is small enough.
• Mar 3rd 2010, 08:05 PM
frenchguy87
Quote:

Originally Posted by pickslides
This is quite a strange function to use with this method as it is fully factored and gives you the roots.

Anyhow you choose the interval $[0,7]$ and find the signs of $f(0) <0$ and $f(7)>0$
Now bisect the interval $\frac{7-0}{2} = 3.5$ to make 2 smaller intervals and find the sign of $f(3.5)$

We know $f(3.5)>0$ so we choose the smaller intervals with opposite signs $f(0) <0$ and $f(3.5)>0$ , bisect again and so on until the interval is small enough.

Ok that's kind of what I thought... and yea it gives you all the roots but they want you to find the root that will be found after using that method, which apparently ends up being $x=1$
• Mar 3rd 2010, 08:26 PM
tn11631
Quote:

Originally Posted by frenchguy87
let $f(x)=(x-1)(x-2)(x-3)(x-4)(x-5)$. This function has five roots on the interval $[0,7]$. I have to use the bisection method on this interval and figure out what root is located.
I'm not entirely sure I know how to use this method correctly..
Here it goes,
First, since $f(0)<0$ and $f(7)>0$, there exists a $c\in\mathbb{R}$ in between those two points such that $f(c) = 0$ ..
Now would you divide the interval in two?

Wow thats crazy i just took an exam today with almost that same exact problem just a little difference in like one of the numbers. Anyway the way I did it was that I just kept applying the bisection method where a=0 b=7 and you have to find thier signs and all that, but then when you plug in p_n it either matches the sign of a or b, however you keep doing it until f(p)=0. However in your case you can kind of see where its going and it keeps getting closer and closer to 1 so your root that will be determined would be 1. I'm not sure if its all the correct but I mean it worked for me on both my hw's and the practice exam. Hope this helped
• Mar 3rd 2010, 09:18 PM
frenchguy87
Quote:

Originally Posted by tn11631
Wow thats crazy i just took an exam today with almost that same exact problem just a little difference in like one of the numbers. Anyway the way I did it was that I just kept applying the bisection method where a=0 b=7 and you have to find thier signs and all that, but then when you plug in p_n it either matches the sign of a or b, however you keep doing it until f(p)=0. However in your case you can kind of see where its going and it keeps getting closer and closer to 1 so your root that will be determined would be 1. I'm not sure if its all the correct but I mean it worked for me on both my hw's and the practice exam. Hope this helped

How would you formally write the end of the proof, whenever you have a small enough interval?
I am down to the $[0,1.75]$ interval, so how would I go one to say that the root must be 1?
• Mar 3rd 2010, 09:36 PM
Drexel28
Quote:

Originally Posted by frenchguy87
How would you formally write the end of the proof, whenever you have a small enough interval?
I am down to the $[0,1.75]$ interval, so how would I go one to say that the root must be 1?

I don't understand. Are you to play dumb? Clearly $f(x)=0,\text{ }x=1,2,3,4,5$ and $f(x)\ne 0\text{ }x\in \mathbb{R}-\{1,2,3,4,5\}$. So clearly the zero given must be one.
• Mar 3rd 2010, 10:06 PM
frenchguy87
Quote:

Originally Posted by Drexel28
I don't understand. Are you to play dumb? Clearly $f(x)=0,\text{ }x=1,2,3,4,5$ and $f(x)\ne 0\text{ }x\in \mathbb{R}-\{1,2,3,4,5\}$. So clearly the zero given must be one.

lol I just wasn't sure how to write it properly... sorry