Using fzero to find the second intercept of the two curves.

So I have function f(x)=-ln(x+0.01)+e^x that plots the cross sectional area of a shape and then a horizontal line segment given by g(x)=f(x)x=0. I need to find the second intercept of the curves. Easy to visual if you graph it. Now I need to use the fzero function to solve this and I have been searching for hours for an example like this and can't find anything similar.

Answer 1

[John D'Errico]

Do you need to provide a start point for fzero? (Yes.)

Does fzero work best if you provide a bracket (two points), that surrounds the root in question? (Yes.)

So if you know roughly where the root of interest is, then what is the problem? You cannot use fzero if you do not provide a start point at least.

Answer 2

[John BG]

Hi Andy

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

With the following you get both zero points of g(x)=f(x)*x=0

1.

1st point with fsolve start point x01=0

close all;clear all;clc;
f1=@(x) log(x+.01)+exp(x)-x*(log(x+.01)+exp(x))
x01=0;
options = optimoptions('fsolve','Display','iter');
options.MaxFunctionEvaluations = 1000 
y1=fsolve(f1,x01,options)

f1 =

    function_handle with value:

      @(x)log(x+.01)+exp(x)-x*(log(x+.01)+exp(x))

options = 

    fsolve options:

     Options used by current Algorithm ('trust-region-dogleg'):
     (Other available algorithms: 'levenberg-marquardt', 'trust-region')

     Set properties:
                       Display: 'iter'
        MaxFunctionEvaluations: 1000

     Default properties:
                     Algorithm: 'trust-region-dogleg'
                CheckGradients: 0
      FiniteDifferenceStepSize: 'sqrt(eps)'
          FiniteDifferenceType: 'forward'
             FunctionTolerance: 1.000000000000000e-06
                 MaxIterations: 400
           OptimalityTolerance: 1.000000000000000e-06
                     OutputFcn: []
                       PlotFcn: []
      SpecifyObjectiveGradient: 0
                 StepTolerance: 1.000000000000000e-06
                      TypicalX: 'ones(numberOfVariables,1)'
                   UseParallel: 0

     Show options not used by current Algorithm ('trust-region-dogleg')

                                           Norm of      First-order   Trust-region
   Iteration  Func-count     f(x)          step         optimality    radius
       0          2         12.9973                           377               1
       1          4         4.02556      0.0344646           49.7               1
       2          6        0.712178      0.0809282            7.6               1
       3          8        0.050945      0.0937519            1.1               1
       4         10     0.000680152       0.046356         0.0991               1
       5         12     2.09695e-07     0.00686238        0.00168               1
       6         14     2.17453e-14    0.000124849       5.41e-07               1

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the default value of the function tolerance, and
the problem appears regular as measured by the gradient.

<stopping criteria details>

y1 =
   0.262487926852786

.

2.

2nd zero point with fzero starting x02=100

x02=100;
y2=fsolve(f1,x02,options)

                                           Norm of      First-order   Trust-region
   Iteration  Func-count     f(x)          step         optimality    radius
       0          2     7.08218e+90                      7.15e+90               1
       1          4     9.58373e+89       0.989999       9.68e+89               1
       2          6     1.29689e+89       0.989899       1.31e+89               1
       3          8     1.75496e+88       0.989797       1.77e+88 

..

     104        210        0.591762        0.36162           3.44            1.34
     105        212       0.0156953       0.171887          0.383            1.34
     106        214     4.06374e-05      0.0410109         0.0175            1.34
     107        216     4.00574e-10     0.00232187       5.46e-05            1.34
     108        218      4.0027e-20    7.33586e-06       5.46e-10            1.34

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the default value of the function tolerance, and
the problem appears regular as measured by the gradient.

<stopping criteria details>

y2 =

     1.000000000073332

.

If you find this answer useful would you please be so kind to consider marking my answer as Accepted Answer?

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thanks in advance

John BG

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