If the square (x2-m) of the function f (x) = (x-1) has a maximum at x = 1, then the value range of M is

If the square (x2-m) of the function f (x) = (x-1) has a maximum at x = 1, then the value range of M is

f(x)=(x-1)^2(x^2-m)
f'(x)=2(x-1)(x^2-m)+2x(x-1)^2=2(x-1)[2x^2-x-m]
f'(1)=0
For the maximum value, we also need to:
f'(1-)>0,
f'(1+)

If the maximum value of function f (x) = - X3 + MX2 + 1 (m ≠ 0) in (0,2) is the maximum value, then the value range of M is___ ．

F ′ (x) = - 3x2 + 2mx = - 3x (x-2m3), Let f ′ (x) = 0, then x = 0 or x = 2m3. And ∵ the maximum value of function f (x) = - X3 + MX2 + 1 (m ≠ 0) in (0, 2) is the maximum value, ∵ 0 ＜ 2m3 ＜ 2, and the function f (x) monotonically increases on (0, 2m3), monotonically decreases on (2m3, 2), and ∵ 0 ＜ m ＜ 3. So the answer is: (0, 3)

Given the function f (x) = sin (KX / 5 + π / 3) (k is not equal to 0), write the maximum value m, minimum value m and minimum period T of F (x); try to find the minimum positive integer k so that
When the independent variable x changes between any two integers, the function f (x) takes m and m at least once

1. When KX / 5 + π / 3 = π / 2 + 2n π (n is an integer), the,
F (x) = 1 is the maximum
2. When KX / 5 + π / 3 = - π / 2 + 2n π (n is an integer), the,
F (x) = - 1 is the minimum
3. The minimum period T = 2 π / (K / 5) = 10 π / K
4. (Please add questions)

The even function f (x) defined on R is an increasing function at [0, ∞). When x 1 and x 2 ∈ [- 3 / 2,3 / 2], we compare the size of F (Tan x 1) and f (Tan x 2)

Because y = TaNx increases in X ∈ [- 3 / 2,3 / 2], y may be less than 0
If | x1 | > | x2 |, f (tanx1) > F (tanx2);
If | x1 | = | x2 |, f (tanx1) = f (tanx2);
If | x1|

The even function f (x) defined on R is an increasing function at (0, positive infinity). X1 and X2 belong to (- 3 / 2,3 / 2). Compare the big difference between F (tanx1) and f (tanx2)
help

If X1 = X2, then f (tanx1) = f (tanx2)
If the absolute value of X1 is less than the absolute value of X2, then f (tanx1) < f (tanx2)
If the absolute value of X1 is greater than the absolute value of X2, then f (tanx1) > F (tanx2)

Given the function F X = a ^ x (a > 0 and a is not equal to 0), if X1 is not equal to X2, prove f ((x1 + x2) / 2)

There are two ways to do this:
One is to use the basic inequality
F (x1) + F (x2) = a ^ X1 + A ^ x2 ≥ 2 times of the root sign (a ^ X1 multiplied by a ^ x2) = 2 times of the root sign (x1 + x2 power of a)
Divide by 2 to get 1 / 2 [f (x1) + F (x2)] ≥ (x1 + x2 power of a)
What is the (x1 + x2 power of a) under the root sign? It is f (sum of X1 and X2 divided by 2) (that is the known one). You see, f (sum of X1 and X2 divided by 2) = a ^ [1 / 2 (x1 + x2)], half of which is the root sign
If x 1 ≠ x 2 is known, the original inequality holds
The second is to use the image of concave function
The left side of the inequality is the midpoint of the curve between two points, and the right side is the midpoint of the line between two points
O(∩_ ∩) oh~

Define the length of the interval | x1, X2 | as x2-x1, the domain of the function f (x) = 3 ^ | x | is [a, b] and the range is [1,9]. Find out the maximum and minimum length of [a, b]?

The minimum is the logarithm of log3 as the base 2, and the maximum is the minimum value of 2

The even function f (x) defined on R satisfies: for any x 1 x 2 belonging to (negative infinity, 0] (x 1 ≠ x 2), if x 2-x 1 / F (x 2) - f (x 1) > 0, then ()
A f（-5）＜f（4）＜f（6） B f（4）＜f（-5）＜f（6）
C f（6）＜f（-5）＜f（4） D f（6）＜f（4）＜f（-5）

All have x2-x1 / F (x2) - f (x1) > 0
We can see that f (x) increases on (negative infinity, 0]
In R, f (x) is an even function, so f (x) decreases on (0, + infinity)
So f (4) > F (5) > F (6)
On R, f (x) is an even function, that is, f (5) = f (- 5)
So f (4) > F (- 5) > F (6)
So choose C

Try to discuss the number of zeros of function y = MX2 + 3x-1

m=0 1
M / = 0 can see what he means

Try to find an interval of length 1 where the function y = (x-1) / (3x + 2) has at least one zero point

Let y = 0
x=1
As long as the interval contains x = 1, it can be used
[a, a + 1] where 0