f(x) = e^x-3x matlab code

I want to write a matlab script that finds solutions of function f(x) = e^x - 3x. I tried to write some code. My matlab code should include iteration table and graphic of function. Can anyone help me please with a code or give me advices?

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 = (a+b)/2;
      if (f(r) == 0)
          break;
      elseif (f(a)*f(r) <= 0)
          b = r ;
      else
              a = r;
      end
  end

[Matt J]

solutions of function f(x) = e^x - 3x

I think what you really mean is, solutions of the equation

   exp(x) - 3*x = 0

[Adomas Bazinys]

It would be also interesting how to find solutions of exp(x)-3*x = 0. How should I change my code so that it could solve this equation?

[Rik]

Did you read Steven Lord's comment? Your current code finds 1 solution to this equation already.

If you have some restriction (e.g. because this is a homework assignment), please mention them.

Answer 1

[John BG]

Hi Adomas Bazinys

this is John BG <mailto:[e-mail] [e-mail]>

What you want to do is, the way you want to do it, indeed good exercise, but

1.

What about fsolve

At Mathworks the function fsolve was time ago developed precisely so that the people using MATLAB could save time avoiding the need for any writing of such functions, and go straight to the roots, have a look:

fun = @(x) exp(x)-3*x;
x0 = [1.5,0];
x = fsolve(fun,x0)

x =
   1.512134551657842   0.619061283355628

2.

Don't start with x0=[0 0] or [1 0] because then fsolve only returns a single real solution

3.

The requested graph

fplot(fun);grid on

.

4.

And the table

x_range1=0;x_range2=10;xrange_step=1;
x1=[x_range1:xrange_step:x_range2]';fx1=fun(x1);

T1=table(x1,fx1)
T1 =
    11×2 table
      x1           fx1        
      __    __________________
       0                     1
       1    -0.281718171540954
       2      1.38905609893065
       3      11.0855369231877
       4      42.5981500331442
       5      133.413159102577
       6      385.428793492735
       7      1075.63315842846
       8      2956.95798704173
       9      8076.08392757538
      10      21996.4657948067

I assumed you ask for a MATLAB table.

You can also export this table to Excel, Word or a simple .txt file, if you choose to call those the intended tables to hold the result.

5.

Checking the 4 limits +- Inf +-1j*Inf

syms n;limit(exp(n)-3*n, n, -inf)
=
Inf
syms n;limit(exp(n)-3*n, n, inf)
=
Inf

syms n;limit(exp(1j*n)-3*1j*n, n, inf)
 =
NaN
syms n;limit(exp(1j*n)-3*1j*n, n, -inf)
 =
NaN

Do you really need considering exp(1j*imx)-3*1j*imx ?

If so the start point is

exp(1j*imx)-3*1j*imx = cos(imx)+1j*(sin(imx)-3*imx)

this would be when real(x)>>imx=imag(x)

6.

the 2 real roots found with fsolve probably turn into a corona of complex roots that may be appreciated developing a bit further the following start point.

     rangelim=15;rangestep=.01;
     rex=[-rangelim:rangestep:rangelim];
     imx=[-rangelim:rangestep:rangelim];

     [RX,IX]=meshgrid(rex,imx);

     Z=exp(RX+1j*IX)-3*(RX+1j*IX);
     figure(2);hs=surf(abs(Z),RX,IX);hs.EdgeColor='none';

Adomas

If interested in the complex roots, please let me know.

If you find the so far supplied answer useful, would you please be so kind to consider marking my answer as Accepted Answer?

To any other reader, if you find this answer useful please consider clicking on the thumbs-up vote link

thanks in advance

John BG

<mailto:[e-mail] [e-mail]>

[Steven Lord]

The fsolve function is part of Optimization Toolbox, but you don't need that toolbox for a one-variable, one-equation problem. The fzero function in MATLAB can handle those problems. Of course I suspect this is a homework assignment and the professor won't accept code that simply calls either fsolve or fzero.

Answer 2

[Rik]

If you want to store intermediate result, us the n for indexing. Also, don't you mean if abs(f(r(n)))<=tol? Then you would actually use the variable tol.

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)')

Answer 3

[maryam]

Find an approximation to within 0.00001 to a value in [0, 1] with 𝑓(𝑥) = 𝑒 𝑥 − 𝑥 2 + 3𝑥 − 2 = 0 Use Bisection method

[Walter Roberson]

No, that will certainly not solve the question posted by Adomas