Need to solve e^x=3*x in tho ways

I need to find an interval, wherein is one solution of given equation. I need to solve equation e^x = 3*x in two ways: using Bisection and Newton methods, so I need two codes. My code should also include iteration table and graphic, in which I could see at least some iterations of solution. I already started to program but I'm absolutely new at Matlab and worked a lot so I'm asking for help. Both methods are written for f(x) = e^x - 3*x, but now I need program for e^x = 3x.

This is in Bisection method.

f=@(x) exp(x) - 3*x ;
a = 0; b = 1;tol = 1e-8;
if (f(a)*f(b) > 0)
    error ('invalid choice of interval')
end
r=0; n=0;
while ((b-a)/2 > 0)
    n = n + 1;
    r(n) = (a+b)/2;%#ok suppress warning, we can't know the length of r in advance
    if (f(r(n)) == 0)
        %if abs(f(r(n)))<=tol
        break;
    elseif (f(a)*f(r(n)) <= 0)
        b = r(n) ;
    else
        a = r(n);
    end
end
it_table=[r' f(r)'];
clc
disp(it_table)
figure(1),clf(1)
plot(1:numel(r),f(r))
xlabel('Iteration number'),ylabel('f(r)')

And this is in Newton method:

   clear all
   close all
   clc
   f=@(x) exp(x) - 3*x % Change here for different functions
   df=@(x) exp(x) - 3 %this is the derivative of the above function
   a=0; b=1;
   x=a;
   for i=1:1:100
   x1=x-(f(x)/df(x));
   x=x1;
   end
   sol=x;
   fprintf('Approximate Root is %.15f',sol)
   a=0;b=1;
   x=a;
   er(5)=0;
   for i=1:1:5
   x1=x-(f(x)/df(x));
   x=x1;
   er(i)=x1-sol;
   end
   plot(er)
   xlabel('Iteration number')
   ylabel('Error')
   title('Error vs iteration number')

Answer 1

[Rik]

The roots of f(x)=exp(x)-3*x are the same as solving exp(x)=3*x.

[Adomas Bazinys]

you mean that f(x)=exp(x)-3*x and exp(x)=3*x is the same??

[Adomas Bazinys]

so, is my Bisection code already gets solutions of exp(x)=3*x?

[Rik]

"you mean that f(x)=exp(x)-3*x and exp(x)=3*x is the same??"

Almost. The zero of the f(x) function is the same as solving exp(x)=3*x for x.

"so, is my Bisection code already gets solutions of exp(x)=3*x?"

Yes, but as John BG mentioned, only 1 solution. To find the other solution, you will have to change your initial guesses.

[Adomas Bazinys]

Ok, thanks. And how can I make my f(x) equal to 0 in both codes? I have heard about fzero function but don't know how to use it.

[Rik]

Generally the documentation of Matlab is really good. Have you tried doc fzero?

[Adomas Bazinys]

Yes, I have. But I have'nt experience with fzero. How could I use fzero for my functions in my codes? Can anyone help me with some code with fzero?

[Torsten]

But you wrote a bisection and a Newton method to find the roots of your functions - what do you need "fzero" for ?

[Adomas Bazinys]

My current methods are written for finding solutions of f(x) = e^x-3*x. So, I want to make my f(x) equal to 0 (f(x)=0). And I'm now searching for a ways to do that.

[Adomas Bazinys]

Sorry, it might be that I'm not understanding something. Some people said that my codes already are able to find solutions of e^x-3*x=0..

[Torsten]

Bisection method and Newton's method are numerical methods to find zeros of a given function. So they have the same purpose as "fzero" has.

Your current methods are written to find solutions of f(x)=e^x-3*x=0.

[Adomas Bazinys]

Thanks, Torsten!