Adding Tolerence to Bisection Code

I am writing a code that needs a tolerance of 0.000005. My function is (currently using equation 3):

 function [out] = f(x)
 %returns roots
 %out=exp(x)-(x^2+4); %Equation 1
 %out=x - 2^(-x); %Equation 2
 out=x^3 - 1.8999*x^2 + 1.5796*x - 2.1195; %Equation 3
 end

I am working of adding another column, f(P), to my code. Here it is:

 clear; clc;
nr=input('Enter Number of Iterations: ');
A=input('Enter Value of A: ');
B=input('Enter Value of B: ');
n=1;
fprintf('\n');
i=(1:nr);

 %loops
 if f(A)*f(B)>0
    fprintf('Error. Values Cannot Be Iterated.');
 else
     %print column header
     fprintf('\n \t');
     fprintf('     A\t\t\t\t     B \t\t\t        P \t\t\t        f(P)');
     fprintf('\n\n');
    while n<=nr
        fprintf('%.0f\t',i(n));
        P=(A+B)/2;
        %Add F=f(P) formulation here? Maybe?
        toprow=[A(n),B(n),P(n)]; %Add F(n)
        fprintf('% 3.6f\t\t\t',toprow);
        if f(A(n))*f(P(n))<0
            A(n+1)=A(n);
            B(n+1)=P(n);
        elseif f(P(n))*f(B(n))<0
            A(n+1)=P(n);
            B(n+1)=B(n);
        end
        fprintf('\n');
        n=n+1;
    end
end

This is the output that I desire:

 The root to the equation f= x^3-1.8999*x^2+1.5796*x-2.1195 =0 on [1,2] was found to be 1.703133 where f(1.703133)=0.000013.

 The 17 iterations for the bisection routine were recorded as: 

 n    a      b      p      f(p) 
 1    1.000000    2.000000    1.500000    -0.649875    
 2    1.500000    2.000000    1.750000    0.185731    
 3    1.500000    1.750000    1.625000    -0.278558    
 4    1.625000    1.750000    1.687500    -0.058767    
 5    1.687500    1.750000    1.718750    0.060302    
 6    1.687500    1.718750    1.703125    -0.000016    
 7    1.703125    1.718750    1.710938    0.029946    
 8    1.703125    1.710938    1.707031    0.014916    
 9    1.703125    1.707031    1.705078    0.007437    
 10    1.703125    1.705078    1.704102    0.003708    
 11    1.703125    1.704102    1.703613    0.001845    
 12    1.703125    1.703613    1.703369    0.000914    
 13    1.703125    1.703369    1.703247    0.000449    
 14    1.703125    1.703247    1.703186    0.000216    
 15    1.703125    1.703186    1.703156    0.000100    
 16    1.703125    1.703156    1.703140    0.000042    
 17    1.703125    1.703140    1.703133    0.000013

Sorry to take up so much space! I just want to be sure that I add enough information in so that I can get a little help. So, basically, I would like to add in a tolerance of 0.000005 to the code, and I would like for the code to stop and print the following type output when it hits the root. I feel like that would include an 'if P==0' statement added also, but I'm not quite sure. Any and all help and pointers is appreciated! Thank you!

[Basil Saeed]

You're saying "when it hits the root". Do you mean that you want your function to stop if it is within 0.000005 of the root even if it reaches that before n iterations? (Keep in mind that you might also not find the correct solution after n iterations. How should you handle cases like that?)

[Bailey Smith]

Yes. And the numbers that will be used do have roots before the 100th iteration, but I suppose I should add an 'else' statement saying that the root has not been found (just in case). Thank you for that advice. I have been working on it and I think I almost have it completely figured out, though. I'm just stuck on printing the equation from the function on the 'The root to the equation...' part.