Help my code won't run for finding the minimum and maximum of a function

The function is p(x) and when you take the derivative, I get the quadratic formula and I'm trying to use it to find critical points. the quadratic I have is a saved code(not in this code I saved it as a function):

%The quadratic formula as a function
function X=quadratic(A,B,C)

    crit_pt1=(-B+sqrt(B^2-4*A*C))/(2*A)
    crit_pt2=(-B-sqrt(B^2-4*A*C))/(2*A)

This is where my program starts:

    a = -1.0;
    b = 2.0;
    c = 1.0;
    d = 1.0;
    e = 1.0;
    f = 1.0;

    p = @(x) c * x.^3 + d * x.^2 + e * x + f;

    quadratic(3 * c, 2 * d, e);

    if isreal(crit_pt1)
      disp('Crit Point 1:');
      disp(crit_pt1);
      disp('p(Crit Point 1):');
      disp(p(crit_pt1));

      disp('Crit Point 2:');
      disp(crit_pt2);
      disp('p(Crit Point 2):');
      disp(p(crit_pt2));
    else
      disp('The critical points are complex.');
    end

    disp('Left Endpoint:');
    disp(a);
    disp('p(Left Endpoint):');
    disp(p(a));

    disp('Right Endpoint:');
    disp(b);
    disp('p(Right Endpoint):');
    disp(p(b));

    delta = (b - a) / 100;
    x = a:delta:b;
    y = p(x);
    plot(x, y)

It says error to many outputs for quadratic and also that critpt_1 and crit_pt_2 not defined even though I have it defined in the saved function for quadratic. Any help?

[Matt J]

Notice how your code is now in a more readable font, distinct from your text. I did that with this button,

and hope you will do the same from now on.

[cakey]

Thank you

Answer 1

[Mischa Kim]

Cakey, critpt_1 and crit_pt_2 are only defined locally, in function quadratic. Also, the function does not seem to be complete, since there are no values assigned to X. Update the function to

 function X = quadratic(A,B,C)
 crit_pt1 = (-B+sqrt(B^2-4*A*C))/(2*A);
 crit_pt2 = (-B-sqrt(B^2-4*A*C))/(2*A);
 X = [crit_pt1; crit_pt2];
 end

and use in the calling function or script something like

 CP = quadratic(3 * c, 2 * d, e);
 crit_pt1 = CP(1);
 crit_pt2 = CP(2);

That way the critical points are assigned within quadratic to the variable X. In the calling function you can then access the values for the critical points.

[cakey]

Wow, i think your suggestions worked! I don't really understand why the syntax I used was wrong. I understand that my quadratic function was incomplete. But the second part about crit_pt is confusing to me. Thank you. More specifically, i would like to know why CP = quadratic(3 * c, 2 * d, e); crit_pt1 = CP(1); crit_pt2 = CP(2); needs the CP= stuff.

[Mischa Kim]

Think of the function quadratic as a sort of a black box (from the perspective of the calling function). The only thing the calling function knows is that when quadratic is called with three input arguments, it returns a 2-by-1 vector whose first ( CP(1) ) and second ( CP(2) ) elements are crit_pt1 and crit_pt2, respectively.

Answer 2

[Matt J]

I assume you really meant this,

function [crit_pt1, crit_pt2]=quadratic(A,B,C)

    crit_pt1=(-B+sqrt(B^2-4*A*C))/(2*A)
    crit_pt2=(-B-sqrt(B^2-4*A*C))/(2*A)

and later

[crit_pt1, crit_pt2]=quadratic(3 * c, 2 * d, e)

A few more miscellaneous tips.

1. Instead of x = a:delta:b, do instead x=linspace(a,b,101)
2. Instead of p = @(x) c * x.^3 + d * x.^2 + e * x + f do polyval([c d e f],x)