Given the function f (x) = m / 2 (x-1) ^ 2-2x + 3 + LNX, constant M ≥ 1 (1), find the monotone decreasing interval of function f (x) (2) When m = 2, let the domain of function g (x) = f (x) - f (2-x) + 3 be D, any x1, X2 ∈ D, and X1 + x2 = 1. It is proved that one of G (x1) + G (x2), G (x1) - G (x2), G (2x1) + G (2x2), G (2x1) - G (2x2) must be a constant (excluding x1, x2) (3) If the tangent of curve C: y = f (x) at point P (1,1) has and only one common point with curve C, find M

Given the function f (x) = m / 2 (x-1) ^ 2-2x + 3 + LNX, constant M ≥ 1 (1), find the monotone decreasing interval of function f (x) (2) When m = 2, let the domain of function g (x) = f (x) - f (2-x) + 3 be D, any x1, X2 ∈ D, and X1 + x2 = 1. It is proved that one of G (x1) + G (x2), G (x1) - G (x2), G (2x1) + G (2x2), G (2x1) - G (2x2) must be a constant (excluding x1, x2) (3) If the tangent of curve C: y = f (x) at point P (1,1) has and only one common point with curve C, find M

1) Derivation:
f‘（x）=(m*(2*x - 2))/2 + 1/x - 2=0
x1=(m + (m^2 + 4)^(1/2) + 2)/(2*m)；
x2=(m - (m^2 + 4)^(1/2) + 2)/(2*m)
Monotone decreasing interval [X2, X1] of F (x)
2)m=2;f(x)=m/2(x-1)^2-2x+3+lnx=(x-1)^2-2x+3+lnx;
g(x)=f(x)-f(2-x)+3=4*x + ln(x) - ln(x - 2) - 9;
It is proved that there must be a constant. It is OK to take it in and simplify it
3) The tangent equation is Y-1 = f '(1) * (x-1);
But curve C, which curve do you mean?

It is proved that on the interval (1, + infinity), the image of function f (x) = 1 / 2x ^ 2 + LNX is always below the function g (x) = 2 / 3x ^ 3 Detailed process

g(x)-f(x)
=(2/3)x^3-(1/2)x^2-lnx
After derivation
=2x^2-x-(1/x)
=x(2x-1)-(1/x)
Because the range is (1, + infinite), so 2x-1 > 1, x > 1, so x (2x-1) > 1, and because (1 / x) < 1, the derivative is > 0, so g (x) - f (x) increases monotonically
g(1)-f(1)=2/3-1/2>0
So in (1, + infinity), f (x)
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Let f (x) = 1 / 3x LNX (x > 0), then y = f (x) A. There are zero points in the interval (1 / E, 1) and (1, e) B. there is a zero point in the interval (1 / E, 1), and there is no zero point in the interval (1, e) C. There is no zero point in the interval (1 / E, 1) and (1, e) D. there is no zero point in the interval (1 / E, 1), and there is a zero point in the interval (1, e) 2、 Given the quadratic function f (x) = ax ^ 2 + BX + C (a ≠ 0) 1. If f (1) = 0, does the function f (x) have other zeros besides 1? If so, ask for it. If not, please explain the reason 2. If X1 < X2, f (x1) ≠ f (x2), it is proved that the equation f (x) = [f (x1) + F (x2)] / 2 must have a real root in the interval (x1, x2)

The zero point is y = 0, which is the intersection point with the X axis. We can see that when x > 0, then f (x) '< 0, so when x > 0, the function is a subtraction function

(2010. Nankai District the second mock exam) set the function f (x) =1 3x LNX (x > 0), then the function y = f (x) () A. In the interval (1 e. 1) there is no zero point in the interval (1, e) B. In the interval (1 e. There are zero points in 1) and no zero points in the interval (1, e) C. In the interval (1 e. There are zero points in (1, e) D. In the interval (1 e. There is no zero point in (1, e)

The derivative of the function is f ′ (x) = 13 − 1x = x − 33x. When f ′ (x) > 0, the solution x ＞ 3, then the function monotonically increases. When f ′ (x) < 0, the solution 0 ＜ x ＜ 3, then the function f (x) in (1E, 1), (1, e) is a decreasing function, ∵ f (1E) = 13 × 1e-ln1e = 13e + 1 ＞

Given the function f (x) = x? + 3x? + 3x, If f (x) image is translated by vector a to get g (x), and G (x) satisfies g (x + 1) + G (- x + 1) = 1, then the coordinates of vector a are

Let a = (a, b)
We can get g (x) = (x-a) ^ 3 + 3 (x-a) ^ 2 + 3 (x-a) + B
Then substitute g (x + 1) + G (- x + 1) = 1
The coefficient of X term on the left is equal to 0, and the constant term is 1
The result a = (2,3 / 2) can be obtained

Find the monotone interval of the function f (x) = LNX - (1 / 3) x + 2 / (3x)

f'(x)=(1/x)－(1/3)－2/(3x²)=[－(x－2)(x－1)]/(3x²)
Then: F (x) decreases in (0,1), increases in (1,2), and decreases in (2, + ∞)

Given the function f (x) = 1 / 2x Λ 2-alnx, find the monotone interval of function f (x), and prove that when x > 1, 1 / 2x Λ 2 + LNX

First question:
∵f（x）＝（1/2）x^2－alnx,　∴f′（x）＝x－a/x＝（x^2－a）/x.
Let f ′ (x) = (x ^ 2-A) / x > 0, then x ^ 2-A > 0, x > 0; or x ^ 2-A < 0, x < 0
ν x ^ 2 > A, x > 0; or x ^ 2 < A, x < 0
Considering the definition domain of the function, we need x > 0.only: x ^ 2 > A, x > 0
When a ≤ 0, X ＞ 0. When a ＞ 0, X ＞ 0
When a ≤ 0, the increasing interval of the function is (0, + ∞), and there is no decreasing interval
When a > 0, the increasing interval of the function is (√ a, + ∞), and the decreasing interval of the function is (0, √ a)
Second question:
Let f (x) = (1 / 2) x ^ 2 + LNX - (2 / 3) x ^ 3
The derivative is obtained as follows: F ′ (x) = x + 1 / x-2x ^ 2, f ″ (x) = 1-1 / x ^ 2-4x
Obviously, when x > 1, f ″ (x) = 1-1 / x ^ 2-4x < 0,
When x ＞ 1, f ′ (x) = x + 1 / x-2x ^ 2 is a decreasing function, while f ′ (1) = 1 + 1-2 = 0,
When x ＞ 1, f ′ (x) ＜ 0, when x ＞ 1, f (x) = (1 / 2) x ^ 2 + LNX - (2 / 3) x ^ 3 is a decreasing function,
F (1) = 1 / 2 + 0 - (2 / 3) = 3 / 6-4 / 6 = - 1 / 6 < 0,
When x > 1, f (x) = (1 / 2) x ^ 2 + LNX - (2 / 3) x ^ 3 < 0,
∴（1/2）x^2＋lnx＜（2/3）x^3 .

Given the function f (x) = 1 / 3x ^ 3-ex ^ 2 + MX + 1 g (x) = LNX / x, find the monotone interval (2) of the function f (x) for any X1 and x2 If G (x)

The derivation of F (x) is obviously a positive quadratic formula, that is, the parabola with the opening upward, and then judge whether there is the possibility of negative value. If there is, in the negative value range of derivative, it is a decreasing function, and the positive range is an increasing function. The key is m value. When m value is in a certain interval, the function increases first, then decreases and then increases
The first question is done, and the second one is very simple,
The train of thought knows, the solution is not difficult! Oneself practice more!

How to understand the concavity and convexity of a curve and how to draw the picture of unknown function according to the derivative?

If f (x) is continuous on [a, b], if V x 1, x 2 ∈ (a, b)

How to judge the image of original function according to the image of derivative function?

It is mainly seen that the derivative function is greater than 0, the original function is increased, the derivative function is less than 0, and the original function is reduced,