The minimum value of function f (x) = 1 / 2x2 LNX

The minimum value of function f (x) = 1 / 2x2 LNX

If f '(x) = X-1 / x = (x + 1) (x-1) / X (the domain of definition is x > 0), Let f' (x) = 0, then x = 1 (- 1 rounding)
When 0 < x < 1, f '(x) < 0, when x > 1, f' (x) > 0, so when x = 1, f (x) is the smallest
So f (x) min = f (1) = 1 / 2
x^3 +6x² +11x+6
a²(a+2b)² - 9(x+y)²
1、f(-3)=0
By the factor theorem, the long division method is used
So = (x + 3) (x ^ 2 + 3x + 2)
=(x+3)(x+1)(x+2)
2、=(a^2+2ab)^2-(3x+3y)^2
=(a^2+2ab+3x+3y)(a^2+2ab-3x-3y)
﹙1﹚x³＋6x²＋11x＋6
＝x³＋6x²＋9x＋2x＋6
＝x﹙x＋3﹚²＋2﹙x＋3﹚
＝﹙x＋3﹚﹙x²＋3x＋2﹚
＝﹙x＋1﹚﹙x＋2﹚﹙x＋3﹚
(2) a and 178; (a + 2b) & 178
﹙1﹚x³＋6x²＋11x＋6
＝x³＋6x²＋9x＋2x＋6
＝x﹙x＋3﹚²＋2﹙x＋3﹚
＝﹙x＋3﹚﹙x²＋3x＋2﹚
＝﹙x＋1﹚﹙x＋2﹚﹙x＋3﹚
﹙2﹚a²﹙a＋2b﹚²－9﹙x＋y﹚²
＝﹙a²＋2ab﹚²－﹙3x＋3y﹚²
= (A & # 178; + 2ab-3x-3y) (A & # 178; + 2Ab + 3x + 3Y)
x³+6x²+11x+6
=（x³+x²）+（5x²+11x+6）
=x²（x+1）+（x+1）（5x+6）
=(x+1)(x²+5x+6)
=(x+1)(x+2)(x+3)
Given function f (x) = LNX + 2A / (x + 1)
When x is greater than 0 and X is not equal to 1, LNX / (x-1) greater than a / (x + 1) is constant, and the value range of a is obtained
1.f'(x)=1/x-2a/(x+1)^2=[x^2+(2-2a)x+1]/[x(x+1)^2],
(1-A) ^ 2-1 = a ^ 2-2aa / (x + 1) is constant
A = 0, f (x) is an increasing function, f (1) = 0,
00, H (x) is an increasing function,
When x → 1, H (x) → LNX + (x + 1) / X (Law of Robida) → 2,
∴a
Decomposition factor: x2-4xy + 4y2-2x + 4y-3
x2-4xy+4y2-2x+4y-3=（x-2y）2-2（x-2y）-3=（x-2y-3）（x-2y+1）．
Given function f (x) = LNX - ax ^ 2 + (2-A) x (A & gt; 0)
Known function f (x) = LNX - ax ^ 2 + (2-A) x (a > 0)
1. Finding monotone interval of F (x)
2. It is proved that when 0 = 0
According to the following questions:
lnx1-ax1^2+(2-a)x1=0.(1)
lnx2-ax2^2+(2-a)x2=0.(2)
f'(xo)=1/xo-2axo+(2-a)>=0.(3)
x1+x2=2xo.(4)
Simultaneous (1) ~ (4) elimination of a
2(x2-x1)/(x1+x2)-ln(x2/x1)>=0
That is, 2 [(x2 / x1) - 1] / [1 + (x2 / x1)] - ln (x2 / x1) > = 0
Note x2 / X1 = t > 1
And the function is introduced
h(t)=2(t-1)/(t+1)-lnt,t>1
Derivation is easy to get
H '(T) = - (t-1) ^ 2 / [t (T + 1) ^ 2] 1, and H (T) can be continuous at t = 1, then
h(t)1
That is, 2 [(x2 / x1) - 1] / [1 + (x2 / x1)] - ln (x2 / x1)
How to extract the common factor by factorization!
I can't learn this just now
Under what circumstances should the number be advanced?
There is also the question 6 (m-n) & #179; - 12 (n-m) & #178; = 6 (m-n) & #179; - 12 [- (m-n)] & #178; = 6 (m-n) & #179; - 12 (m-n) & #178; = 6 (m-n) & #178; (m-n-2)
I can't understand how, (m-n-2) how did this come about?
And square
1 .(m+n)²-n²
2.49(a-b)²-16(a+b)²
3.(2x+y)²-(x+2y)²
4 .(x²+y²)²-x²y²
5... 3ax & # 178; - 3ay quartic power
6... P quartic-1
Given the function f (x) = - a2x2 + ax + LNX (a is a real number), if the function f (x) is a decreasing function on (1, positive infinity), find the range of A
The derivative is f '(x) = - 2A ^ 2 * x + A + 1 / X
f ’（x）1
Substitute x = 1 so that f '(1) = - 2A ^ 2 + A + 1
Extraction of common factors
1. Quadratic-2a + B of (2a-b)
2. The shape of one wall of a house is composed of a rectangle and a triangle. If the wall is designed into a square shape, the area remains unchanged, and the length of the bottom edge is still a, what is the height
The length of the bottom edge of the rectangle is a, the height is B, and the height of the triangle is h
1. Quadratic-2a + B of (2a-b)
=(2a-b)²-(2a-b)
=(2a-b)(2a-b-1)
2. The height is the height of the rectangle + half of the height of the triangle
Given the function f (x) = (x2-3x + 2) LNX + 2009x-2010, the interval where f (x) must have zero is ()
A. （0，1）B. （1，2）C. （2，3）D. （2，4）
F (1) = - 1 ＜ 0, f (2) = 2008 ＞ 0, f (1) f (2) ＜ 0, and the interval where f (x) must have zero is (1,2)
The following factors are decomposed by extracting common factors
(1) The fourth power of 3x & # 179; + 6x
(2)6p(p+q)-4q(p+q)
3x³+6x⁴
=x³(1+2x)
6p(p+q)-4q(p+q)
=(6p-4q)(p+q)
=2(3p-2q)(p+q)
Hello, landlord!!
To extract the common factor is to extract the long image
The first question I think about is to mention something related to X when it looks like it
The second question P + Q looks like me
It's much easier to do that
Answer 3x & # 179; (1 + 2x) 2 (3p-2q) (P + Q)
I wish you progress in your study!!