Need help looking for local minimum on a parabola

I was wondering if anyone would be able to help me out with this.. I need to use a function M file to determine the local minimum of a parabola where the critical points can also qualify as a minimum. Here is my code. Currently it is returning values without error but the values don't match any of my test data given by my instructor.

if true
  function c = myminimum(f, a, b)
syms x v w;
w=solve(diff(f(x)));
v=subs(f(x), w);
if (a<v && a<b);
  c=a;
elseif (b<v && b<a);
  c=b;  
else c=v;   
end
end

Here is the test data if that helps:

if truea = myminimum(@(x) x^2+1,-3,2)
a = 1
a = myminimum(@(x) x^2+6*x-3,-2,0)
a = -11
a = myminimum(@(x) -2*x^2+10*x,3,8)
a = -48
a = myminimum(@(x) -x^2+2*x+1,5,6)
a = -23

Answer 1

[Star Strider]

Do you have to use the if block?

The easy way is to use the min function:

syms x v w;
w=solve(diff(f(x)));
z = [w a b];
v=subs(f(x), z);
[q,k] = min(v);
c = v(k);

If you must use the if loop:

syms x v w;
w=solve(diff(f(x)));
z = [w a b];
v=subs(f(x), z);
if (v(1)<v(2) && v(1)<v(3));
  c=v(1);
elseif (v(2)<v(3));
  c=v(2);  
else
    c=v(3);
end

If it makes you feel any better, the ‘correct’ answer to this set:

a = myminimum(@(x) x^2+6*x-3,-2,0)
a = -11

is in error. The correct answer is -12, corresponding to the minimum of the parabola.

[Rascad]

Ahh I see.. Yeah the if loop is required. I did get -12 for that one at some point but the rest were different than the given data. Thank you for your help!

[Star Strider]

My pleasure!

The rest of your professor’s answers were correct. Only that one wasn’t.

[Justin Wyss-Gallifent]

Prof here - the -11 was correct, the student is not understanding (or not asking) the problem correctly.

[Star Strider]

Interesting. I got the impression from the way Rascad asked the question that it was a sort of ‘multiple choice’ between the minimum of the function or the two provided points. Nothing in the Question mentioned that the minimum was to be on the interval (a,b). In that instance of course -11 is correct because -3 is not on the interval (-2,0).

[Justin Wyss-Gallifent]

Yes, he/she didn't ask the question clearly (or perhaps misunderstood, since I'm getting the same confusion from other students). The actual project question asks for the minimum on the interval [a,b] and even makes a comment that this minimum may occur at the critical point if the critical point is in the interval and may occur at an endpoint.

[Star Strider]

Open interval as well. I’ll keep that in mind if I see other questions. It would be nice if your students uploaded a PDF of (or a URL to) the problem statements, possibly with some background and your preferences as to the limits of our help, so that those of us here could offer appropriate help rather than amplify the obfuscation.

[Dan]

I'm having trouble with this same problem and tried using both the min function code and the if loop code you wrote, Star Strider, but I keep getting errors and I don't know how to fix them.

These are the errors I get from the code with the if loop: Error using sym/subsindex (line 663) Invalid indexing or function definition. When defining a function, ensure that the body of the function is a SYM object. When indexing, the input must be numeric, logical or ':'.

Error in sym>privformat (line 1473) x = subsindex(x)+1;

Error in sym/subsref (line 685) [inds{k},refs{k}] = privformat(inds{k});

Error in myminimum (line 3) w=solve(diff(f(x)));

[Star Strider]

See if my comments here help.