Can someone check my code?

I am suppose to create a function m file called myminimum(f,a,b) which takes three inputs:

f: A function handle.

a: A real number.

b: A real number.

Note: You may assume that f(x) is a quadratic function and that a < b.

Does: Finds the minimum value of f(x) on [a, b] keeping in mind that this may occur either at the

critical point (if the critical point is in [a, b]) or at one of the endpoints. You’ll probably

want to use some if statements to check cases.

Returns: The minimum value of f(x) on [a, b].

Here is my code:

function m=myminimum (f,a,b);
syms x;
y=subs(f(x),a);
z=subs(f(x),b);
w=subs(diff(f(x)),0);
m=min(y,z,w);
end

It keeps returning an error. The error says "MIN with two matrices to compare and a working dimension is not supported." Can someone tell me how to fix my code?

Here is some sample data and the correct answers to the same code.

a= 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

Answer 1

[Star Strider]

I refer you to: Need help looking for local minimum on a parabola.

[Bri]

Not to be rude, but I was reading the comments on that question and just wanted to know if you modified the code at all you posted originally on that questions?

[Star Strider]

I haven’t modified it since I posted the Answer (and had a brief conversation with your professor).

Note that the Original Poster of that Question didn’t mention that you have to determine the minimum on the open interval [a,b].

So you need to do both:

• Determine the value of f(x) at the critical point and the limits of the interval,
• Determine that the critical point is on the interval, since if it is not, do not consider the value of f(x) at the critical point in determining the minimum.

So the only part of your code you still need to do is to determine that the critical point is on the interval, and then output (as m) the minimum of f(x) on the interval. Note that those tests are not part of my posted code, because your classmate didn’t mention they were necessary.

In the code you posted, you still need to define the derivative, solve for the inflection point, and find f(x) at the inflection point before you evaluate f(x) at the interval limits:

syms x
f = sym(f);
df = diff(f);
cp = solve(df);
w = subs(f,cp);

After that, you need to determine if ‘cp’ is on the interval. If it is, take the minimum of all three points: f(a), f(b), f(cp), else consider f(a) and f(b) only.

[Bri]

Can you please check my code now? It will not return a value when I run

(@(x) 1*x^2+7*x+8, -0.200000, 6.800000)

Can you tell me what I did wrong now?

function m=myminimum (f,a,b);
syms x;
y=subs(f(x),a);
z=subs(f(x),b);
derv=diff(f(x);
c=solve(c==0,x);
w=subs(f(x),c);
if(c>=a & c<=b & w<y & w<z);
    m=vpa(w);
elseif (y<z && y<w);
    m=vpa(y);
elseif(z<y && z<w);
    m=vpa(z);
end
end

[Star Strider]

I put the extra vpa calls in for my convenience. You can take them out if you wish.

The problem is that once the first (initial) if condition is satisfied, you can break out of the block because you know the critical point is on the interval and the value of the function there is the minimum. After that, all you need to do is to test to see if y<z.

This works:

syms x;
y=vpa(subs(f(x),a));
z=vpa(subs(f(x),b));
derv=diff(f(x));
c=vpa(solve(derv==0,x));
w=vpa(subs(f(x),c));
if(c>=a && c<=b && w<y && w<z);
    m=vpa(w);
    break
elseif (y<z);
    m=vpa(y);
else
    m=vpa(z);
end

[Bri]

Thank you so much, you have been very helpful throughout the duration of my matlab class. I really appreciate it.

[Star Strider]

I took the essence of your code outside of the function (exactly as I posted it) because I run everything I test on MATLAB Answers in a test script file that doesn’t allow function definitions, so break worked. Replace break with return and all should be well.

[Star Strider]

My pleasure!

Answer 2

[siham boujrad]

%This program computes to calculate the Total Resistance (or Impedance) for Series or Parallel Circuit.

R1=input('Enter the value of R1'); %ohms

R2=input('Enter the value of R2'); %ohms

R3=input('Enter the value of R3'); %ohms

R=[R1,R2,R3];

RTs=sum(R);

RTp=1/sum(1./R);

If RTs>=5 & RTs<25

fprintf('Resistors are in series and R_Total = %7.2f ohms \n',RTs)

elseif RTp>0 & RTp<=1

fprintf('Resistors are in parallel and R_Total = %7.2f ohms \n',RTp)

else disp('Error')

endif

Answer 3

[siham boujrad]

plz i need someone to check the code for me. Thank you