fminbnd question, program doesn't work?

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Joseph Percsy
Joseph Percsy el 5 de Abr. de 2013
This is what I have so far.
clear
x = -2:.01:4;
y = x.^2-2*x-4;
y=originlen(x);
plot(x,y)
pause % Press any key to continue
x1=input('Enter left hand endpoint of subinterval: ');
x2=input('Enter right hand endpoint of subinterval: ');
xmin = fminbnd('originlen',x1,x2)
ymin=originlen(xmin);
originlen is this
function d=originlen(x)
x=-2:0.01:4;
d=sqrt(x.^2+(x.^2-2*x-4).^2);
plot(x,d)
This is currently the error that I'm getting:
??? Operands to the and && operators must be convertible to logical scalar values.
Error in ==> fminbnd at 260 if ( (abs(p)<abs(0.5*q*r)) && (p>q*(a-xf)) && (p<q*(b-xf)) )
Error in ==> plotting at 12 xmin = fminbnd('originlen',x1,x2)
What I want the program to do is display the point on the parabola that is closest to the origin, (found by d(x)), and the minimum distance.

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