Asked by Eralp Ali Zeydan
on 12 Jun 2019

What is the maximum value assumed by the function f(x) = sin(5*cos(x))*cos(5*sin(x)) on the interval [0,1] ?

Answer by James Browne
on 12 Jun 2019

Greetings,

I wrote an example for you which finds the maximum value of f(x) over the specified interval and plots the resuts. Here you go, hope this helps!

%Define the interval over which f(x) will be evaluated

xInterval = [0 1];

%specify resolution of the evaluation

dx = 0.0001;

%Create a vector of x values based on the interval of evaluateion and

%resolution specifications

x = xInterval(1): dx : xInterval(2);

%Calculate f(x) values

for i = 1:length(x)

fx(i) = sin(5*cos(x(i)))*cos(5*sin(x(i)));

end

%Find the magnitude and index of the maximum f(x) value and print the

%maximum magnitude to the command window

[M,idx] = max(fx);

fprintf('The maximum value in the given range of x is: %5.4f\n',M)

%Use the index of the maximum f(x) value to find the x value which produced

%the maximum f(x) value

xForMaxFx = x(idx);

%Determine title based on interval of evaluation parameters

titleName = strcat('Maximum f(x) Value over the interval [',num2str(xInterval(1)),',',num2str(xInterval(2)),']');

%Plot f(x) and highlight the maximum value

plot(x,fx,xForMaxFx,M,'kd')

title(titleName)

xlabel('x (units)')

ylabel('f(x) (units)')

ylim([-1,1])

legend('f(x)','Maximum f(x) value','location','southeast')

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Answer by Walter Roberson
on 12 Jun 2019

Step 1: Plot the function to see visually the approximate peak. Find an approximate interval for the peak; it does not have to be precise at all.

Step 2: Differentiate the function. Solve for a zero of that over the approximate interval that you identified. That gives the location of the peak.

Step 3: substitute the location of the peak into the formula to get the value of the peak.

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## Walter Roberson (view profile)

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https://ch.mathworks.com/matlabcentral/answers/466761-the-maximum-value-assumed-by-the-function-on-the-interval#comment_714140

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