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

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Adomas Bazinys
on 25 Feb 2018

Commented: Walter Roberson
on 25 Oct 2022

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

##### 3 Comments

Rik
on 25 Feb 2018

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.

### Accepted Answer

John BG
on 25 Feb 2018

Hi Adomas Bazinys

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

##### 1 Comment

Steven Lord
on 25 Feb 2018

### More Answers (2)

Rik
on 25 Feb 2018

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)')

##### 0 Comments

maryam
on 25 Oct 2022

##### 1 Comment

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