# 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
Rik on 25 Feb 2018
If you have some restriction (e.g. because this is a homework assignment), please mention them.

John BG on 25 Feb 2018
What you want to do is, the way you want to do it, indeed good exercise, but
1.
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';
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?
John BG
Steven Lord on 25 Feb 2018
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.

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)')

maryam on 25 Oct 2022
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 on 25 Oct 2022
No, that will certainly not solve the question posted by Adomas