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Can the monotonicity of derivative function deduce the monotonicity of function
Not at all. For example, a simple quadratic culvert number
ax＾2+bx+c
It is definitely not a monotone implication number, but its derivative is a primary function
ax+b
How to find the derivative of distribution function and probability density of standard normal distribution? I made a mistake in this formula, which means how to get the right of the equal sign after deriving from the left of the equal sign?
I happened to do this when I went to self-study this evening. I thought about it for half an hour before I came to my bedroom to figure out what was going on Φ' (x)= φ (x) , you take the derivative of the left form directly and get - 4 / A ^ 2* φ' (2 √ Y / a), and because φ (x) = 1 / √ 2 π * e ^ - x ^ 2 / 2 is the probability density of the standard normal distribution. Are you right φ (x) When you take the derivative, you'll find φ' (x)=(-x)* φ (x) , the left formula = (- 4 / A ^ 2) * (- 2 √ Y / a) can be obtained by substituting x = 2 √ Y / A* φ (x)=(8√y/a^3)* φ (2 √ Y / a) = right type
The problem of solving the second derivative of an implicit function differential X ^ 2 + 3Y ^ 2 = 5 find the second derivative, i.e. d2y / DX2 (2 is the lower case 2, in the upper position) I can only find the first dy / DX = - X / 3Y, and then how to find the derivative of this equation?
d2y/dx2=-1/3*[Y-Xdy/dx]/Y^2=-(y+x^2/3y)/3y^2
It is known that f (x) is an increasing function defined on (0, positive infinity) and satisfies f (XY) = f (x) + F (y), f (2) = 1. If x satisfies f (x) - f (X-2) greater than 3, f(xy)=f(x)+f(y),f(2)=1, ∴f(8)=3f(2)=3, f(x)-f(x-2)>3, F (x) > F (8) + F (X-2) = f (8x-16), F (x) is an increasing function defined on (0, positive infinity), ∴x>8x-16>0, The solution is 2F (8) + F (X-2) = f (8x-16), How? Can you tell me what you know? I'm not good at this class
You calculate f (8) = 3f (2) = 3 above, so the inequality f (x) - f (X-2) > 3 is reduced to f (x) - f (X-2) > F (8), then you get: F (x) > F (8) + F (X-2)
And because the condition is: F (XY) = f (x) + F (y), f (8) + F (X-2) = f [8 (X-2)] = f (8x-16)
See?
If the function FX satisfies: for all real numbers x and y, FX + FY = x (2y-1) holds (1) find F0
If x = 0, y = 0, then F0 + F0 = 0, 2f0 = 0, F0 = 0. If x = 1, y = 0, then F1 + F0 = - 1, F1 = - 1
Let the definition field of the function be (0, + ∞), and f (XY) = f (x) + F (y) is constant for any positive real number x, y Let the definition domain of the function be (0, + ∞), when x > 1, f (x) < 0, and f (XY) = f (x) + F (y) is constant for any positive real number x, y, f (2) = 1, solve the inequality f (x) + F (X-2) < 3
There seems to be a mistake in the title. When x > 1, f (x) < 0, and then f (2) = 1 comes out, isn't it self contradictory
RELATED INFORMATIONS
- 1. Let the domain of the function y = f (x) be (0, + infinity), and f (XY) = f (x) + F (y) be constant for any positive real number x, y, and f (2) = 1, and 1. Find f (8) 2. Solve inequality f (x) + F (X-2) < = 3
- 2. Fxy = FX + FY, F1 / 3 = 1, if FX + f2-x < 2, find the value range of X Let y = FX be the subtractive function over the definition field R
- 3. It is known that the function FX is meaningful for x > 0 and satisfies F2 = 1, fxy = FX + FY, and FX is an increasing function. If FX + F (X-2) = 2 is true, the value range of X is
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