Bisectional method Not always giving correct roots

My code sometimes does not give correct roots? Sometimes it will and other times it wont depending on the boundaries, please let me know if you guys know what is causing it.

clc

clear all
close all

f=@(x)(x.^3+7*x.^2-33*x-135);
yl = input('type approximated lower boundary:');
yu = input('type approximated upper boundary:');
err=input('Type desired approximate error limit:');
   
if(f(yl)*f(yu)) > 0 
   disp('TRY NEW BOUNDARIES');
  return
   %if values do not produce a negative number give better boundaries
end

while abs(yu-yl) >= err %basically how close do you want the boundaries to go
    ynew=(yl+yu)/2;
    if (f(yl)*f(yu) < 0)
        yu=ynew;
    else
        yl=ynew;
  
    
    end
end
fprintf('The root of this equation given your error limit is=%f', ynew);

Answer 1

[Geoff Hayes]

Anthony - can you provide an example (with inputs) that gives the correct solution and an example that doesn't? Also, don't you need to include the new value, ynew, in your comparison

    ynew = (yl + yu) / 2;
    if (f(yl) * f(ynew)) < 0
        %etc.
    end

rather than re-using the upper bound again as

    if (f(yl)*f(yu) < 0) % incorrect?

[Anthony Ming]

Something that doesnt work:22,-30,.001.

One that works is 18,-6,.001.

I tried this after putting in the correction (I believe youre right) But it's still throwing bad answers:/)

[Anthony Ming]

It seems to give correct answers more often now.. Maybe it has something to do with the checker to see if you can use those boundaries. It will still display an answer when it gives of an error if you cant use those boundaries.

[Geoff Hayes]

Anthony - I don't understand your inputs: 22,-30,.001. Isn't the first input supposed to be the lower bound on the interval and the second input the upper bound? So shouldn't the order be reversed to -30, 22, 0.001?

[Geoff Hayes]

For your function, there seem to be roots at (at least) -9 and 5. Is this true?

As for your second input set, what happens if you try entering in the order -6, 18, 0.001?

[Anthony Ming]

yes you are correct, I just typed them in wrong.

Here:

-20,20,.0001 gives 9.999 (Root is actually -9,-3,5)

one that is correct is:(-600,300,.001)

Thank you

[Anthony Ming]

Yes, -9,-3, and 5. When I type those inputs you gave it does give an answer however my code says "Ignore this answer" I put that because im not sure how to keep it from displaying.

But it will still throw bad answers when they are within acceptable boundaries

[Geoff Hayes]

You may also want to consider

while abs(yu-yl) >= err

Do we need another condition to determine if we are finished? Take a look at Bisection Method algorithm. What other check do they have?

[Anthony Ming]

ah I got it! It should be while its less than or equal to. It's working properly.

Thank you so much!!

Can you tell me why it threw off numbers? Im guessing it did an extra iteration due to it not being less than which threw it off?

Again, thank you so much

[Anthony Ming]

It actually needed the <= on the statement below the if statement. Otherwise it didnt work.

[Geoff Hayes]

I think that the equality is needed to handle the case where f(ynew) is zero, like in the article with the If f(c) = 0.