Errors when using fminbnd function

Hi, I am trying to find use the fminbnd function to find the minimum and maximum values of a function.

I am getting these errors when I try and use the fminbnd function in my code:

Error using fcnchk (line 106): FUN must be a function, a valid character vector expression, or an inline function object

and

Error in fminbnd (line 194): funfcn = fcnchk(funfcn,length(varagin));

Here is my code

x=-10:.0001:10

f = (2+(x-1.45).^2)./(3+3.5.*(.8.*x.^2-.6.*x+2))

minusf=-1.*f

[xmin,y]=fminbnd(f,-10,10)     *(THIS IS THE LINE THAT THE ERROR APPEARS AT)*

[xmax,y2]=fminbnd(minusf,-10,10)

fmin=f(xmin)

fmax=f(xmax)

[James Robinson]

Any help will be greatly appreciated!! Thanks in advance!

Answer 1

[Walter Roberson]

You have a discrete system. You can just search for its extremes directly.

x=-10:.0001:10
f = (2+(x-1.45).^2)./(3+3.5.*(.8.*x.^2-.6.*x+2))
[fmin, minidx] = min(f);
xmin = x(minidx);
[fmax, maxidx] = max(f);
xmax = x(maxidx);

If you want to use a continuous system then:

f = @(x) (2+(x-1.45).^2)./(3+3.5.*(.8.*x.^2-.6.*x+2));
[xmin, fmin] = fminbnd(f, [-10 10]);
[xmax, fmax] = fminbnd(@(x) -f(x), [10 10]);

[James Robinson]

I am required to use fminbnd, and when I tried your second suggestion there, it didn't work. It popped up the error message:

Error using fminbnd (line 98) FMINBND requires three input arguments.

Thanks for your help

[Walter Roberson]

f = @(x) (2+(x-1.45).^2)./(3+3.5.*(.8.*x.^2-.6.*x+2));
[xmin, fmin] = fminbnd(f, -10, 10);
[xmax, fmax] = fminbnd(@(x) -f(x), -10, 10);
fmax = -fmax;

[James Robinson]

That works! Thanks so much!