Monotone decreasing interval of function f (x) = square of x-2|x-1|

Monotone decreasing interval of function f (x) = square of x-2|x-1|

f(x)={x^2-2x+2 (x≥1）
{x^2+2x-2 (x
Since f (2x + 1) = x2-2x, it is found that f (- 3) - Middle East is 2x + 1 and - 3 are enclosed in brackets
We can calculate x = - 2 by doubling + 1 = - 3
If f (2 × + 1) = x2-2x, X is - 2
Let f (- 3) = 4 + 4-1 = 7
When x > 1, f (x) = x ^ 2-2x-2, f (x) is derived, f '(x) = 2x-2 > 0
When x0, when x
f（x）=｛1.（x+1）² （x≤-1） 2.2x+2 （-1＜x＜1） 3.（1/x）-1 （x≥1）｝.
If f (m) > 1, the range of M is obtained
[the bracket after the formula is the range]
Consider it in three paragraphs,
1. M ≤ - 1 and (M + 1) ^ 2 > 1 -- > m ≤ - 1 and M0 -- > m
What is the monotone decreasing interval of the square of the function y = - 3 4x-x?
If y decreases, then 3 ^ (3x-x & # 178; 0 increases
3 ^ x is an increasing function
So it's 3x-x & # increasing
3x-x²
=-(x-3/2)²+9/4
Opening down
So x
y=-(x+17)^2-289
So the monotonic decreasing interval is [- 17, + ∞)
Is the function y = - 3 + 4x-x ^ 2?
If yes, then y = - (X-2) ^ 2 + 1,
The opening of the parabola is downward, and the axis of symmetry x = 2,
Therefore, the monotone decreasing interval is: [2, + ∞).
When the function f (x) = - X & # 178; + 2x + 3 is an increasing function in the interval (+ infinity, M], find the value of the real number M
The axis of symmetry is x = 1 and the opening is downward,
In order to make the function be an increasing function on (- ∞, M],
The interval needs to be on the left side of the axis of symmetry,
So what is the range of M
m∈(-∞,1]
Let f (x) = X3 power - 3x square - 9x + 1, (1) find the monotone increasing interval of F (x), (2) find the maximum value of F (x) in the interval [- 2,2]
This part is not very familiar
1) From F '(x) = 3x & # 178; - 6x-9 = 3 (X & # 178; - 2x-3) = 3 (x-3) (x + 1) = 0, the extreme point x = 3, - 1 is obtained
The monotone increasing interval is: x > 3, or X
Given the function f (x) = 2x & # 178; + 3, if f (a) = 1, then a=_____
2A ^ 2 + 3 = 1 A ^ 2 = - 1 a = I, I is an imaginary unit! You should be a junior high school student! Then there is no solution. When you go to senior high school, there will be solution!
The solution is 2A & # 178; + 3 = 1, so a & # 178; = - 1, so a is equal to the complex number I
A = 1I I = - 1 imaginary number
The decreasing interval of function y = x2-6x is ()
A. （-∞，2]B. [2，+∞）C. （-∞，3]D. [3，+∞）
The equation of symmetry axis of ∵ function y = x2-6x is x = 3, and the corresponding image is a parabola with the opening upward. As shown in the figure, the subtraction interval of ∵ function y = x2-6x is (- ∞, 3]
The minimum positive period of function f (x) = cos & # 178; 2x
f(x)=cos²2x=(1+cos4x)/2=cos4x/2+1/2.
The period of cos4x is (2 π / 4 =) π / 2, so the period of F (x) is π / 2
The monotone decreasing interval of the function y = 4sin (1 / 3x + π / 4) is
Process is the best, 3Q
π/2+2kπ≤1/3x+π/4≤3/2π+2kπ
3π/4+6kπ≤x≤15π/4+6kπ
The function y = f (x) is an even function on (- ∞, + ∞). When x ≥ 0, y (x) = x & # 178; - 2x-3, find the analytic expression of the function y = f (x)
When x0,
Thus, if f (x) is an even function, then
When x