The definition field of MX square - 6mx + m + 8 under the function y = root is r, and find the range of M

The definition field of MX square - 6mx + m + 8 under the function y = root is r, and find the range of M

y=√(mx ²- The domain of 6mx + m + 8) is r
When m = 0 √ 8 = 2 √ 2 > 0, it meets the meaning of the question
When m > 0, △ = 36m ²- 4m(m+8)<=0
The solution is 0, so m range [0,1]

How do I integrate partial derivatives? Find the general integral form of the following formula,, the little girl is really out of her mind ∂(r∂t/∂r)/∂r=Mr ; Where m is a constant

∂(r∂t/∂r)/∂r=Mr ; Where m is a constant, t is a function of what variable? For example, t (R, s)
R ∂ T / ∂ r = Mr ^ 2 / 2 + C (s). When r > 0, ∂ T / ∂ r = Mr / 2 + C (s) / R,
T = Mr ^ 2 / 4 + C (s) LNR + D (s), where C and D are functions of other variables s
Just a little thought,

Find the extreme value of the square of the binary function z = x + the square of Y - 2x + 2Y + 2

Z=x^2+y^2-2x+2y+2=(x-1)^2+(y+1)^2+4≥4
When x = 1, y = - 1

What is the extreme value of binary function f (x, y) = x square + y square + 2XY

Min = 0. No max

F (x, y) = square of X + square of Y + XY, extreme value under condition x + 2Y = 4

According to x + 2Y = 4, so x = 4-2y. It is brought into the original equation. So f (x, y) = (4-2y) ^ 2 + y ^ 2 + (4-2y) y = 3Y ^ 2-12y + 16 = 3 (Y-2) ^ 2 + 4, so when y = 2, it is the minimum. At this time, x = 0

Find the total differential of the function z = f (x, y) determined by equation 2xy-3xyz + ln (XYZ) = 0

By differentiating 2xy-3xyz + ln (XYZ) = 0, we can get 2ydx + 2xdy-3 (yzdx + xzdy + xydz) + (1 / XYZ) (yzdx + xzdy + xydz) = 0 ‰ DZ = (2ydx + 2xdy-3yzdx-3xzdy + DX / x + dy / y) / (3xy-1 / z), which can be simplified by ourselves. It's too troublesome to type on the Internet. We can simplify the fraction and the formula of DX and Dy