MATLAB: How to make a function that returns the difference between the passed function’s max and min value given a range of the independent variable

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I was given the task to develop a function M-file that returns the difference between the passed function's maximum and minimum value given a range of the independent variable. In addition, I have to make the function generate a plot of the function for range. I was given three instances to test my M-file.

a.) f(t) = 8e^(-0.25t)sin(t – 2) , from t=0 to 6pi

b.) f(x) = e^(4x)sin(1/x) , from x =0.01 to 0.02

c.) The built in humps function from x = 0 to 2

I found the exact problem on Chegg, but it provides absolutely no explanation of the code and I would really like to be able to understand this problem. I hope someone can help!

Best Answer

Hint: You need to use * for when you multiply, and use exp() for e^, and you can use linspace() to get a bunch of values for t between some starting and stopping values.

t=linspace(0,6*pi, 1000); % Get 1000 values of t
f = 8 * exp(-0.25*t).*sin(t - 2);
plot(t, f);
grid on;

I'm assuming you know how to use min() and max() and how to subtract the values they return.

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MATLAB: Substitute variable and find maximum of the function

The only function provided by Mathworks, of which I am aware, which can efficiently handle such mixed integer nonlinear programming problems is ga(): https://www.mathworks.com/help/gads/ga.html. However, you need to have global optimization toolbox to use this function

fun = @(x, q) (sin(x(1)).^2 + log(x(2)))./(x(1) + q(1) + q(2));
lb = [1;1;0;0]; % lower bounds
ub = [5;5;1;1]; % upper bounds
[sol, f] = ga(@(x) -fun(x(1:2), x(3:4)), 4, [], [], [], [], lb, ub, [], [3 4]);

Result:

>> sol
sol =
    1.0000    5.0000         0         0
>> f
f =
   -2.3175

MATLAB: Getting error when asking from max and min for negative values.

Sounds like you may be trying to index into the array using the min or max value instead of the index:

mn = min(data);
minValue = data(mn);

Don't do that. Use both values if you want to get the index. A better way:

[minValue, indexOfMin] = min(data);
[maxValue, indexOfMax] = max(data);

Or you may have used min for the variable

min = min(data);  % Horrible idea

in which case when you came along and tried to do it again, it would have used your min array (a vector is data was a 2-D array) as indexes into your newly created min array. min() is no longer the min function if you replaced it with a variable of the same name. So if the min was now an array of column mins because you stored the mins in an array called min, then doing min(data) will throw that error if any elements of data have negative, zero, or floating points (fractional) values.

Best Answer

Related Solutions

MATLAB: Substitute variable and find maximum of the function

MATLAB: Getting error when asking from max and min for negative values.

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