How to find the derivative method of monotonicity

It is not the derivative method of monotonicity, but the derivative method to judge the monotonicity of function,
For the given function y = f (x), if y 'is greater than 0, it is a monotone increasing function and if y' is less than 0, it is a monotone decreasing function

The extremum of function y = x ^ 3-3x ^ 2-9x-5 is judged by the second derivative

y=x^3-3x^2-9x-5
y'=3x^2-6x-9
Let y '= 0, that is, x ^ 2-2x-3 = 0
The solution is X1 = - 1, X2 = 3
y''=6x-6
Y '' | (x = - 1) = - 120, x = 3 is the minimum point
The minimum value is 27-27-27-5 = - 34

It is known that the function f (x) has the second derivative in a certain field of 1, f (x)'monotonically decreases, and f (1) = f '(1) = 1
Why is it wrong?
My idea is this:
Although the derivative function at point 1 is decreasing, it is at least greater than 0, so f (x) is monotonically increasing in the field of 1. So f (x) > x in the left field of 1, and f (x) in the right field of 1

Let f (x) = f (x) - X & nbsp; & nbsp; derivative be f '(x) - 1 when x = 1 & nbsp; f' (1) = 1 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; X & gt; 1 & nbsp; f '(1) & gt; 1 & nbsp; & nbsp; & nbsp; & nbsp; f' (x) & gt; 0 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; when X & lt

Find the monotone interval and extremum of y = x ^ 3-3x ^ 2-9x + 5?

Using the method of derivation
The derivative of y = x ^ 3-3x ^ 2-9x + 5 is
y'=3x^2-6x-9
Let y '= 3x ^ 2-6x-9 = 0
x=-1 x=3
Therefore, two extreme points - 1 and 3 are obtained
When x = - 1, y = 10
When x = 3, y = - 22
So the function y = x ^ 3-3x ^ 2-9x + 5
When x = - 1, the maximum value is 10, and when x = 3, the minimum value is - 22
The monotone interval is as follows
Monotonically increasing on (- infinity, - 1), (3, + infinity)
Monotonically decreasing on [- 1,3]

According to the geometric meaning of complex addition, it is proved that ||||||||||||||||||||||||||||||||||||

|Z1 |, | Z2 |, | Z1 + Z2 | form three sides of a triangle (it can be a degenerate triangle). According to the sum of the two sides is greater than or equal to the third side, the difference between the two sides is less than or equal to the third side

Let the complex numbers Z1 and Z2 satisfy | Z1 | = | Z2 | = 2, | Z1 + Z2 | = 2 √ 3, then | z1-z2 | = 3?

Let Z1 = a + biz2 = C + di. According to the meaning of a ^ 2 + B ^ 2 = 4C ^ 2 + D ^ 2 = 4 (a + C) ^ 2 + (B + D) ^ 2 = (2 √ 3) ^ 2 = 12, the third formula subtracts the previous two formulas and deduces 2Ac + 2bd = 4| Z1 - Z2 | ^ 2 = (A-C) ^ 2 + (B-D) ^ 2 = a ^ 2 + B ^ 2 + C ^ 2 + D ^ 2 - (2Ac +

The problem of solving complex number I
(1-i)^6/(1+i)^5

If the fraction is multiplied by (1-I) ^ 5, I write it separately for observation
It's (1-I) ^ 11,
The following is (1 + I) ^ 5 * (1-I) ^ 5
The following calculation is 2 ^ 5,
The final result is (1-I) ^ 11 / 32
When the process is (1 + I) ^ 5 * (1-I) ^ 5, (1 + I) * (1-I) = 1 -- 1 = 2, so (1 + I) ^ 5 * (1-I) ^ 5 = 32

It is known that the equation 2x ^ 2-2 (1 + I) x + ab - (a-b) I = 0 has real roots (A.B belongs to R)
Find the value range of A.B
The range of real roots

Let the real root be t. then (2t ^ 2-2t + AB) + (b-a-2t) I = 0, so there must be 2T ^ 2-2t + AB = 0, b-a-2t = 0. Substitute 2T = B-A into the first formula, (B-A) ^ 2 / 2 - (B-A) + AB = 0, that is, a ^ 2 + B ^ 2 + 2a-2b = 0

Plural questions,
If ω = - 1 / 2 + root 3 / 2 * I
Calculation ((root 3 + I) / 2) ^ 6 + (- root 3 + I / 2) ^ 6
The original formula is (- root 3 conjugate ω - I) ^ 6 + (root 3 ω - I) ^ 6
I don't know what to do next

You are wrong. The original formula ((√ 3 I + 1) / 2) ^ 6 + (- √ 3 I + 1) / 2) ^ 6 should be changed into
(√ 3 ω) ^ 6 + (√ 3 ω) ^ 6
=27 conjugate ω ^ 6 + 27 ω ^ 6
=27 (conjugate ω ^ 6 + ω ^ 6)
So the original formula is 27

How to find the derivative method of monotonicity