Let f (x). G (x) monotonically increase in the interval (a, b). It is proved that the functions ψ (x) = max {f (x), G (x)} and ω (x) = min {f (x), G (x)} also increase in (a, b)

Let f (x). G (x) monotonically increase in the interval (a, b). It is proved that the functions ψ (x) = max {f (x), G (x)} and ω (x) = min {f (x), G (x)} also increase in (a, b)

The known function f (x) = 2Sin (ω x + φ) (ω > 0, │φ

If t / 2 = 6-2, t = 8, then w = 2pai / T = Pai / 4
f(2)=2sin(Pai/4*2+@)=2
sin(Pai/2+@)=1
|@|

High school mathematics derivative proof function f (x) > G (x) proof f (x) min > G (x) max, can do so?

No, because x with the minimum value of F is not necessarily equal to x with the maximum value of G

Given the function f (x) = 6-x2, G (x) = x, define f (x) = min (f (x), G (x)), then f (x) max=

2. Draw an image and find the intersection point
The intersection of the first quadrant is obtained

Let f (x) = min {x + 2,4-x}, then f (x) max

For this kind of nested maximum value problem, we should first understand the meaning of the problem: whenever x takes a number, x + 2 and 4-x have a smaller number, and the problem is to find the maximum value of the smaller number. The following three methods are given: 1. Drawing method. First, draw y = x + 2 and y = 4-x in the plane rectangular coordinate system, and they have an intersection (1,3)

Find the gradient gradf | m of function f = ln (X & # 178; + Y & # 178; + Z & # 178;) at point m (1,2, - 2)

Partial derivation first
AF / AX = 2x / (x ^ 2 + y ^ 2 + Z ^ 2) substitute the point into = 2 / 9
AF / ay = 2Y / (x ^ 2 + y ^ 2 + Z ^ 2) substitute the point into = 4 / 9
AF / AZ = 2Z / (x ^ 2 + y ^ 2 + Z ^ 2) substitute the point into = - 4 / 9
gradf|M=(2/9)i+(4/9)j+(-4/9)k

Let u = ln (x2 + Y2 + Z2), then gradu | (P) =?

∂U/∂x = 2x / (x^2+y^2+z^2)
∂U/∂y = 2y / (x^2+y^2+z^2)
∂U/∂z = 2z / (x^2+y^2+z^2)
grad U = 2(xi + yj + zk) / (x^2+y^2+z^2)
grad U (1,1,1)= 2 (i + j + k) / 3
That's 2 / 3 R, where R is a vector

Find the directional derivative of the function z = ln (x ^ 2, y ^ 2) along the gradient direction at point (3,4)

Y = ln (LNX), find (d2y / DX2) | x = E2, find the second derivative of the function at the specified point!

dy/dx＝1/lnx * 1/x＝1/(xlnx)
d2y/dx2＝-1/(xlnx)^2 * (lnx+x*1/x)＝-1/(xlnx)^2 * (lnx+1)＝- (lnx+1)/(xlnx)^2
Substitute x = e ^ 2 to get [(d2y / DX2) | x = e ^ 2] = - 3 / (2e ^ 2) ^ 2 = - 3 / 4 * e ^ (- 4)

From the equation x ^ 2Y ^ 3 + 4x ^ 2Y ^ 2-9 = 0, y is determined to be the implicit function of X, and = y '?

Two sides of X are derived
2xy^3+x^2*3y^2*y'+8xy^2+8x^2y*y'=0
We get: y '= - (2XY ^ 3 + 8xy ^ 2) / (3x ^ 2Y ^ 2 + 8x ^ 2Y) = - 2XY ^ 2 (y + 4) / [x ^ 2Y (3Y + 8)] = - 2Y (y + 4) / [x (3Y + 8)]