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Finding the extremum of function y = 2x + 8 / X
When
x> 0, 2x + 8 / x > = 2 √ (2x * (8 / x)) = 8
x
8. Let f (x, y) = (1 / 2) x ^ 2 + X + (1 / 3) y ^ 3-y, find the extremum of the function
Let f (x, y) = (1 / 2) x ^ 2 + X + (1 / 3) y ^ 3-y
fx=x＋1=0
fy=y²-1=0
X = - 1, y = 1 or - 1
therefore
The stagnation point is (- 1, - 1) (- 1,1)
fxx=1,fxy=0,fyy=2y
1.（-1,-1）
A=1,B=0,C=-2
AC-B & # 178; < 0, not extreme point
2.（-1,1）
A=1,B=0,C=2
AC-B & # 178; > 0, and a > 0
therefore
The minimum f (- 1,1) = 1 / 2-1 + 1 / 3-1 = - 7 / 6
The extremum of the function z = f (x, y) determined by the following equation x ^ 2 + y ^ 2 + Z ^ 2-8xz-z + 8 = 0
It means that functions with XYZ can't do it, and functions with XY can do it,
1) First find the stationary point, which is the derivative of X on both sides of the implicit function: 2x + 2zz'x-8z-8xz'x-z'x = 0, get: z'x = (8z-2x) / (2z-8x-1), and the derivative of Y on both sides: 2Y + 2zz'y-8xz'y-z'y = 0, get: z'y = 2Y / (- 2Z + 8x + 1) let z'x = z'y = 0, get: 8z-2x = 0, 2Y = 0, substitute y = 0, x = 4Z into the original equation, get: 16Z ^ 2 + Z ^
Finding the extremum of function z = x ^ 2 + (Y-1) ^ 2
z'x=2x z'y=2(y-1)
Let z'x = 0, then x = 0
Let z'y = 0 be y = 1
A=z''xx=2 B=z''xy=0 C=z''yy=2
B^2-AC0
The minimum value of Z = x ^ 2 + (Y-1) ^ 2 at (0,1) is 0
Find the extremum of function z = (x ^ 2 + y ^ 2) ^ (1 / 2) at (0,0)
F (x, y) = (x ^ 2 + y ^ 2) ^ (1 / 2) the image is an inverted cone, which is not differentiable at (0,0) and has no extremum
The extremum and extremum of function f (x) = 3x ^ 4-8x ^ 3 + 6x ^ 2,
f'(x)=3*4x³-8*3x²+6*2x=12x³-24x²+12x
Let f '(x) = 0
Then 12x (x-1) ² = 0
therefore
X = 0, f (x) is increasing
So x = 0 is a minimum
The minimum is f (0) = 0
Y = 8x ^ 3-12x ^ 2 + 6x + 1 do the graph of this function
The second derivative method
Y '= 24x ^ 2-24x + 6 (slope at this point)
Y "= 48x-24 (rate of change of slope at this point)
They can be drawn by the method of point drawing
1. Find the monotone interval and extremum of the function y = 2x to the power of 3-6x to the power of 2 - (8x-1)
If your question (8x-1) is a minus sign, it's easy to do. If it's 8 times the negative power of X, it's more difficult. I'll say it's easy to do. Y = 2x & # 179; - 6x & # 178; - 8x + 1, y '= 6x & # 178; - 12x-8, let y' = 0, х 3x & # 178; - 6x-4 = 0, х x = (3 ± √ 21) / 3
When deriving the implicit function determined by the equation, should the denominator be considered as 0? Why? For example, z = Z ((x, y) is determined by 2x ^ 2 + 2Y ^ 2 + Z ^ 2 + 8xz-z + 8 = 0
That is to say: the partial derivative of Z to X is - (4x + 8Z) / 2Z + 8x-1, where 2Z + 8x-1 is equal to 0?
Not equal to 0
I'll do it all over again, deriving from both ends
4x+2zz'+8z+8xz'-z'=0
z'(2z+8x-1)+4x+8z=0
If 2Z + 8x-1 = 0
At this time, Z 'is meaningless, that is, it does not exist
If 2Z + 8x-1 = 0, a point system (the solution of two ternary equations is a general curve) is obtained
(note that it is not simultaneous with 4x + 8Z = 0, because when 2Z + 8x-1 = 0, it has no meaning for the derivation on both sides of the equation, so 4x + 8Z = 0 has no meaning, and has nothing to do with 2Z + 8x-1 = 0.)
That is, when 2Z + 8x - 1 = 0, the function has no partial derivative on these point systems
So for the partial derivative, 2Z + 8x-1 ≠ 0
Finding the extremum of function f (x, y) = x ^ 2 + y ^ 2-2x + 1
Because f (x, y) = x & # 178; + Y & # 178; - 2x + 1
It can be reduced to two complete squares, namely
f(x,y)=(x-1)²+y²
And because two perfect squares are greater than or equal to zero
So this function has a minimum value equal to 0
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