How would I plot this and then achieve minimum and maximum.
이전 댓글 표시
This is the equation
My script looks like this and fails
x=0:0.1:15
y=(x)^3/3-11(x)^2/2+24x
plot(x,y)
Not yet got my head around MatlB
댓글 수: 1
Dyuman Joshi
2023년 11월 21일
이동: Dyuman Joshi
2023년 11월 21일
Look into
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추가 답변 (1개)
Hi @Tom
The cubic function does not have global extrema, but it does have a local maximum point and a local minimum point. Are you looking to plot something similar to this?
%% Find local max and local min over a certain range
y = @(x) x.^3/3 - 11*x.^2/2 + 24*x;
xlocal = linspace(1, 10, 90001);
ymax = max(y(xlocal))
idx1 = find(y(xlocal) == ymax);
xmax = xlocal(idx1)
ymin = min(y(xlocal))
idx2 = find(y(xlocal) == ymin);
xmin = xlocal(idx2)
%% Plot function over a predefined range
x = 0:0.1:15;
plot(x, y(x)), grid on, hold on
plot(xmax, ymax, 'o', 'markersize', 12, 'linewidth', 2)
plot(xmin, ymin, 'o', 'markersize', 12, 'linewidth', 2)
xlabel('x'), ylabel('y')
xt = [xmax-1 xmin-1];
yt = [ymax+15 ymin+15];
str = {'local max','local min'};
text(xt, yt, str)
댓글 수: 2
Dyuman Joshi
2023년 11월 21일
I'm a curious as to why you chose to use max() and min() instead of islocalmax() and islocalmin().
Sam Chak
2023년 11월 22일
Sometimes, I'm accustomed to using functions with the same name in other software. It's good that you introduced islocalmax() and islocalmin().
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