# How do I find the maximum and minimum of a function in a given domain?

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Ria Singh 2021년 7월 18일
댓글: Rik 2021년 7월 23일
I'm trying to find the max and min of a function over a function, but I can't seem to figure out how. My equation is y = (1*x^4)/4+(4*x^3)/3- 5*(x^2)/2 over -3<=x<=3. I tried doing min(y) and max(y) but it's not working. Does anybody know how to find the max and min???
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Rik 2021년 7월 23일
Original post (in case Ria decides to edit it away again):
How do I find the maximum and minimum of a function in a given domain?
I'm trying to find the max and min of a function over a function, but I can't seem to figure out how. My equation is y = (1*x^4)/4+(4*x^3)/3- 5*(x^2)/2 over -3<=x<=3. I tried doing min(y) and max(y) but it's not working. Does anybody know how to find the max and min???

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### 답변(3개)

Rik 2021년 7월 18일
You need a function like fminbnd:
y =@(x) (1*x.^4)/4+(4*x.^3)/3- 5*(x.^2)/2;
x_min = fminbnd(y,-3,3)
x_min = -2.9999
Let's confirm this with a plot:
fplot(y,[-3 3])
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Image Analyst 2021년 7월 18일
Try this:
x = linspace(-3, 3, 1000);
y = (1*x.^4)/4+(4*x.^3)/3- 5*(x.^2)/2;
plot(x, y, 'b-', 'LineWidth', 2);
grid on;
% Find where min is
[yMin, indexOfMin] = min(y);
fprintf('Min of y at x = %f, y = %f.\n', x(indexOfMin), min(y));
You get
Min of y at x = -3.000000, y = -38.250000.
Is that what you were looking for?
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Walter Roberson 2021년 7월 19일
syms x
y = (1*x.^4)/4+(4*x.^3)/3- 5*(x.^2)/2
y =
LB = -3; UB = 3;
xcrit = solve(diff(y, x),x)
xcrit =
xcrit(xcrit < LB | xcrit > UB) = [];
xcrit = unique([xcrit; LB; UB])
xcrit =
ycrit = subs(y,x,xcrit)
ycrit =
[miny, minidx] = min(ycrit)
miny =
minidx = 1
[maxy, maxidx] = max(ycrit)
maxy =
maxidx = 4
fprintf('minimum is %g at %g\n', miny, xcrit(minidx))
minimum is -38.25 at -3
fprintf('maximum is %g at %g\n', maxy, xcrit(maxidx))
maximum is 33.75 at 3

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