Find the definition field and value field of the following functions (1) y = SiNx + 3 / SiNx + 2 (2) y = under the root sign (1 + SiNx)

Find the definition field and value field of the following functions (1) y = SiNx + 3 / SiNx + 2 (2) y = under the root sign (1 + SiNx)

(1) The domain is positive and negative infinity
The value range is [2 / 3,4]
(2) The domain is positive and negative infinity
The value field is [0, root 2]

Find the conditional extreme value of function u = x + y + Z under condition 1 / x + 1 / y + 1 / z = 1 (x, y, z > 0). Just tell me why the result is "minimum"! other

u=u(1/x+1/y+1/z)=(x+y+z)(1/x+1/y+1/z)=3+x/y+y/x+x/z+z/x+y/z+z/y
Using mean inequality X / y + Y / X ≥ 2 x / Z + Z / X ≥ 2 y / Z + Z / Y ≥ 2

Find the function u = XYZ under additional condition 1 x+1 y+1 z＝1 Extreme value under a (x > 0, Y > 0, z > 0, a > 0)

Finding conditional extremum of multivariate function by Lagrange multiplier method. F (x, y, Z; λ)＝ lnx+lny+lnz− λ (1x+1y+1z−1a)Fx＝1x+ λ 1x2＝0，Fy＝1y+ λ 1y2＝0，Fz＝1z+ λ 1z2＝0 λ＝ − 3a, x = y = z = 3A, the minimum value is 27a3. (3a, 3a, 3a) is the function u = XYZ in addition

In MATLAB, the function f (x, y) = y ^ 3 / 9 + 3 * x ^ 2 * y + 9 * x ^ 2 + y ^ 2 + X * y + 9, - 2

[x,y]=meshgrid(linspace(-2,1,30),linspace(-7,1,30));
f=y.^3/9+3*x.^2.*y+9.*x.^2+y.^2+x.*y+9;
surf(x,y,f);
MAX=imregionalmax(f);
for i=1:1:30
for j=1:1:30
if(MAX(i,j)==1)
hold on;
plot3(x(i,j),y(i,j),f(i,j),'r*');
Text (x (I, J), y (I, J), f (I, J), 'maximum point');
end
end
end

Find the extreme value of function f (x, y) = (x2 + Y2) 2-2 (x2-y2) fast

F (x, y) = (x ^ 2 + y ^ 2) ^ 2-2 (x ^ 2-y ^ 2) FX (x, y) = 4x (x ^ 2 + y ^ 2) - 4x = 4x (x ^ 2 + y ^ 2-1) FY (x, y) = 4Y (x ^ 2 + y ^ 2) + 4Y = 4Y (x ^ 2 + y ^ 2 + 1) stagnation point (0) FXX (x, y) = 4 (x ^ 2 + y ^ 2-1) + 8x ^ 2 fxy (x, y) = 8xyfyy (x, y) = 4 (x ^ 2 + y ^ 2 + 1) + 8y ^ 2ac-b ^ 2

Find the extreme value of binary function f (x, y) = X2 (2 + Y2) + ylny

(1) All stationary points of the function are solved simultaneously by the first derivative = 0
From FX '(x, y) = 2x (2 + Y2) = 0, FY' (x, y) = 2x2y + LNY + 1 = 0
x＝0，y＝1
e．
(2) Using the judgment theorem of extreme value of binary function, the judgment point (0,1
e)   Is it an extreme point
Since f ″ XX = 2 (2 + Y2), f ″ YY = 2x2 + 1
y，f″xy＝4xy，
Set X = 0 and y = 1
e   Bring it in,
f″xx|(0，1
e)＝2(2+1
e2)
f″xy|(0，1
e)＝0
f″yy|(0，1
e)＝e
Because f ″ XX ＞ 0   And (f ″ XY) 2 − f ″ xxf ″ YY ＜ 0, so point (0,1
e)   Is the minimum point of the function
Thus, the binary function has a minimum f (0,1)
e)＝−1
e