clc, clear
figure(), hold on
axis ([-10 10 -10 20])
fplot(@(x) x.^2, [-pi 1],'.-','MarkerSize',12)
hold on, fplot(@(x) x.^3 - 5, [1 2],'.-','MarkerSize',12)
fplot(@(x) 5 - 2*x, [2 pi],'.-','MarkerSize',12)
% Graph formatting
title('Graph of Piecewise Discontinuous f(x)')
legend('y = x^2', 'y = x^3-5','y = 5-2x')
ax = gca;
ax.XAxisLocation = 'origin';
ax.YAxisLocation = 'origin';
grid on
If you need to you can also return the x and y values from fplot using [x,y] = fplot(___)
Another way you could do it:
clc, clear
x = linspace(-pi, pi);
% index for intervals
idxA = x<=1;
idxB = x>1 & x<=2;
idxC = x>2;
% solve equations
y = [x(idxA).^2, x(idxB).^3-5, 5-2*x(idxC)];
figure(), hold on
axis ([-10 10 -10 20])
plot(x(idxA), y(idxA),'.-','MarkerSize',12)
plot(x(idxB), y(idxB),'.-','MarkerSize',12)
plot(x(idxC), y(idxC), '.-','MarkerSize',12)
title(' Graph of Piecewise Discontinuous f(x)')
legend('y = x^2', 'y = x^3-5','y = 5-2x')
ax = gca;
ax.XAxisLocation = 'origin';
ax.YAxisLocation = 'origin';
grid on
% Alternatively
figure(), plot(x,y,'-','LineWidth',2)
axis ([-10 10 -10 20])
set(gca,'XAxisLocation','origin','YAxisLocation','origin')
for reference you can do the loop as follows. You were almost there, but the mistake you made was not to index the current element and also not indexing where to store the output values within y. Compare the code you have to this and you should get the idea.
for ii = 1:numel(x)
if x(ii) <= 1
y(ii) = x(ii)^2;
elseif x(ii) <= 2
disp(1)
y(ii) = x(ii).^3 - 5;
else
y(ii) = 5 - 2*x(ii);
end
end
