MATLAB: Need help plotting a function in matlab

functionhomeworksin

Hi friends, I'm struggling to make a code for above problem. I would appreciate your help.

Best Answer

x = -10:0.05:10;

fun_x_1 = exp(-0.3*x(x<0));

fun_x_2 = sin(x(x>=0).^3)-sin(x(x>=0).^2)+1;

plot(x(x<0),fun_x_1,x(x>=0),fun_x_2)

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MATLAB: Plot with input arguments

funs_cell = {'sin', 'cos', 'tan', 'sinh'};
markers = {'+', 'o', '*', 'x', 's'};
funs = cellfun(@str2fun, funs_cell, 'uniform', 0);
for K = 1 : length(funs)
    plot(x, funs{K}(x), markers{K}, 'DisplayName', funs_cell{K});
    hold on
end

MATLAB: Newton’s Method in Matlab

First, consider if you are trying to solve the wrong problem.

In your question, you state it as e^2*sin(x) - x = 0

However, in your code, you write exp(2*sin(x)) - x. You do realize there is a difference? What I don't know is if you have miswritten your question, or is it your code?

fun = @(x) exp(2)*sin(x) - x;
fplot(fun,[0,3])
yline(0);
fzero(fun,3)
ans =
       2.7589

So the function you claim to want to solve has a zero around x==2.76, which is inconsistent with your claim of where the root lies. Next, look at the function you wrote code for:

fun = @(x) exp(2*sin(x)) - x;
fplot(fun,[-5,3])
yline(0);

Indeed, this does seem to have a root near 2.6357. So, just possibly, you really do want to solve the problem exp(2*sin(x))-x==0.

fzero(fun,3)
ans =
       2.6357

But now, look at the plot! What happens when you start Newton's method at x==1? THINK! Where will the first iteration go? On which side of that hump is x==1?

The point is, Newton's method tries to drive the functino to zero. But if you start the iterations BELOW x==1.5028, which direction will Newton's method push you?

fminbnd(@(x) -fun(x),1,3)
ans =
       1.5028

Think about the meaning of the iterations of Newton's method. What is the goal? What will happen?

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MATLAB: Plot with input arguments

MATLAB: Newton’s Method in Matlab

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