Given that the function f (x) = x3-ax2 + 2ax-1 is an increasing function in the interval (1, + ∞), find the value range of real number a?

Given that the function f (x) = x3-ax2 + 2ax-1 is an increasing function in the interval (1, + ∞), find the value range of real number a?

A:
f(x)=x³-ax²+2ax-1
Derivation: F '(x) = 3x & # 178; - 2aX + 2A
The opening is upward, and the axis of symmetry x = A / 3
F (x) is an increasing function in the interval (1, + ∞)
Then f '(x) > = 0 holds when x > 1
So:
The symmetry axis X = A / 3 = 0 is constant, the discriminant = (- 2A) &# 178; - 4 * 3 * 2A

The function y = f (x) is an odd number on R when x

Right? F (x) = (3x / 9x + 1) - 1 / 2
F (x) = (3x / 9x + 1) - 1 / 2 = [(3x + 1 / 3-1 / 3) / 9x + 1] - 1 / 2 = 1 / 3 - [1 / (27x + 3)] - 1 / 2 = - [1 / (27x + 3)] - 1 / 6

Let x, y ∈ R + and 1x + 9y = 1, then the minimum value of X + y is______ ．

∵ 1x + 9y = 1, x, y ∈ R +, x + y = (x + y) · (1x + 9y) = x + YX + 9 (x + y) y = 10 + YX + 9xy ≥ 10 + 2yx · 9xy = 16 (take "=" if and only if YX = 9xy, x = 4, y = 12)

It is known that M is the point on the ellipse x ^ 2 / 9 + y ^ 2 / 4 = 1, F1 and F2 are the two focal points, I is the inner part of the triangle mf1f2, extending Mi to intersect F1F2 with m, then mi / in =?

It's not convenient to draw a picture, so I use words to describe it
Drawing, heart, angle bisector
So MF1 / MF2 = f1n / f2n
MI/NI=MF1/NF1=MF2/NF2=(MF1+MF2)/(F2N+F1N)=2a/2c=
a/c=3/5^1/2=(3*5^1/2)/5

All prime numbers except 2 and 3 are divided by 6 and the remainder is 1 or 5
Is that 5 prime?

All prime numbers except 2 and 3 are right if they are divided by 6 and the remainder is 1 or 5
5 is also prime, 5 divided by 6, quotient 0 more than 5, in line with the above
The remainder of any number divided by 6 can only be one of 0,1,2,3,4,5; the remainder of prime number divided by 6 will not be 0,2,3,4, otherwise it is not prime number

Given that the function f (x) = x & # 179; + BX & # 178; + CX + D is a decreasing function in the interval [- 1,2], then the maximum value of B + C () a is - 1
Given that the function f (x) = x & # 179; + BX & # 178; + CX + D is a decreasing function in the interval [- 1,2], then B + C ()
A max - 15 / 2 B max 15 / 2
Cmin 15 / 2 dmin-15 / 2

a

Given the function f (x) = x ^ 2 (1 / (2 ^ x-1) + 1 / 2), then f (LG (Lg3)) + F (LG (log3 10)) =?

F (x) = x & sup2; [1 / (2 ^ x-1) + 1 / 2] = x & sup2; [(2 + 2 ^ x-1) / 2 (2 ^ x-1)] = x & sup2; [(2 ^ x + 1) / 2 (2 ^ x-1)] f (- x) = x & sup2; [(2 ^ x + 1) / 2 (2 ^ x-1)] = x & sup2; [(2 ^ x + 1) / 2 (1-2 ^ x)] = - f (x), so it is an odd function LG (log3 10) = LG (LG10 / Lg3) = LG (1 / Lg3) = - LG (Lg3

Fill in the brackets with the appropriate prime number. 20 = () + () = () + () = () + () + ()

20=3+17=7+13=2+7+11

Let the joint probability density of two-dimensional random variables (x, y) be f (x, y) = KX 0

∫∫f（x,y）dxdy=∫kxdx(0-->1)∫dy(0--->x)=∫kx^2dx(0-->1)=k/3=1--->k=3
The edge probability density of X FX (x) = ∫ 3xdy (0 -- & gt; x) = 3x ^ 2 & nbsp; & nbsp; & nbsp; & nbsp; the edge probability density of Y FY (y) = & nbsp; ∫ 3xdx (Y -- & gt; 1) = 3 (y ^ 2-1) / 2

Y = f (2x-1) is even function, x = 0 is symmetry axis, why

Wrong, the axis of symmetry is not x = 0, but x = - 1
Analysis: because y = f (2x-1) is even function
So f (2x-1) = f (- 2x-1)
Then the axis of symmetry is x = (2x-1-2x-1) / 2 = - 1