"Index exceeds array bounds" error; can't resolve

I'm making a function to find min and max of a function (cubic) between a set interval; I keep trying to run my test cases but run into "index exceeds array bounds" in my line 11 (j = g(2)). Some of the individual cases run fine, but it seems that some don't (noticeably the one with a interval of [0, 1.7] which may be a clue.

function rslt = polyMinMax(a,b,c,d,e,f)
  % We have defined rslt to be the final result with the min and max
  % values both for x and y

    polyMinMax1 = [c, d, e, f];
    % This step we take the derivative of the polynomial and get
    % polyMinMax' (prime) which is g

    g = polyder(polyMinMax1);
    h = g(1);
    j = g(2);
    k = g(3);
    % equivalent to a,b,c in normal quad formula

    % This step we look for the solutions given the m parameter
    m = j^2-4*h*k;
    % the discriminant; v1 is positive; v2 is negative

    if m > 0 
        if j >= 0
            v1 = (2*k)/(-j-sqrt(m)); 
            v2 = (-j-sqrt(m))/(2*h);
        else
            v1 = (-j-sqrt(m))/(2*h);
            v2 = (2*k)/(-j+sqrt(m));
        end
    elseif m == 0
            v1 = (-j)/(2*h);
            v2 = NaN;
    else
        v1 = NaN;
        v2 = NaN;
    end

    q = [v1, v2, a, b];

        x1 = min(q);
        x2 = max(q);
        y1 = c*(x1)^3+d*(x1)^2+e*(x1)+f;
        y2 = c*(x2)^3+d*(x2)^2+e*(x2)+f;

        rslt = [x1, x2, y1, y2];

Any ideas? I was thinking it had to do something with j and k being cell arrays and h being a normal character array (since the error starts with j and k and never has been h). I put my test cases below.

-->
rslt = zeros(8,4);
rslt(1,:) = polyMinMax(-1,2,-1,2,-1,1);
rslt(2,:) = polyMinMax( 1,2,1,-2,-1,1);
rslt(3,:) = polyMinMax(-2,1,4,8,-4,-2);
rslt(4,:) = polyMinMax(-1,2,1,0,1,-3);
rslt(5,:) = polyMinMax(-0.3,0.6,1.0e-14,9,-3,0);
rslt(6,:) = polyMinMax(-1,2,0,0,0,1.7);
rslt(7,:) = polyMinMax(0,3,-3,9,-1.0e-14,0);
rslt(8,:) = polyMinMax(0,1,0,-2,3,-1);
fprintf(1, '%11.6g %11.6g %11.6g %11.6g\n', rslt');

Answer 1

[Walter Roberson]

You take the derivative of polyMinMax1 = [c, d, e, f] . If c is non-zero then you would expect to get back a vector of three results, [3*c, 2*d, e] . But if c is zero then you will not get back [0, 2*d, e] : leading zeros are omitted, so you would get back [2*d, e], a vector of length 2. If c and d are both zero, then you would get back [e], a vector of length 1.

rslt(6,:) = polyMinMax(-1,2,0,0,0,1.7);

Right there, c and d are both 0.

[Vishal Sethi]

Ah I see what you mean; makes sense. Thanks! Now I need to go about finding a soln! :)

[Walter Roberson]

LastN = @(M, N) M(end-N+1:end);
ZPadN = @(M, N) LastN( [zeros(1, N), M], N);

g = ZPadN( polyder(polyMinMax1), 3);

[Vishal Sethi]

Endless thanks! I was trying to use if statements to work it but couldn't get it. Infinite thanks.