# Thread: How do I choose the correct range to find zeros

1. ## How do I choose the correct range to find zeros

How do I choose the correct range of numbers to find the zeros...

2. Originally Posted by coopsterdude
How do I choose the correct range of numbers to find the zeros...
Trial and error, We know from Descartes rule of signs that there is exactly one positive root, and 0 or 2 negative roots. Also we know that for large positive x f(x) is positive, and "large" negative x f(x) is negative from the sign of the x^3 term.

So evaluate f(0) it should be -ve, take positive steps untill f changes sign (in this case between 1 and 2). Then as the sum of the roots is equal to -the coefficient of x^2, we conclude that the negative roots if they exist, are >=-5.

There are almost certainly other methods of establishing a suitable range.

RonL

3. A simple way, if you are allowed to use it, is to graph the function (as on a calculator or a computer). Then you can "eye-ball" where the solutions are and use a table to compute them to get the exact values.

-Dan

4. Thanks for your help. I used the trial and error method but it's so time consuming.

Thanks again,

coopsterdude

5. Originally Posted by coopsterdude
Thanks for your help. I used the trial and error method but it's so time consuming.

Thanks again,

coopsterdude
A rather simple method (that computer programs use) is the "bisection method".

First find an interval that has only one zero and work in that interval.

Example,

$y=x^3+x-1$
We can to find the zero of this. That is to solve,
$x^3+x-1=0$
You can draw a graph and see where it intersects only once and work in that interval. Look below.
So work on interval,
$[0,2]$

Now, chose the midpoint $x=1$ does it work?
$1^3+1-1=1>0$ too large.
Meaning the solution must be in the interval.
$[0,1]$.
Now chose the midpoint there $x=.5$ does it work?
$(.5)^3+(.5)-1=-.375$ too little.
Meaning the solution must be in the interval.
$[.5,1]$
Choose the midpoint there $x=.75$ does it work?
$(.75)^3+(.75)-1=.1719$ too large.
Thus, the solution is on the interval
$[.75,1]$
And so on.