Given the function f (x) = - | x | + 1, if the equation f ^ 2 (x) + (2m-1) f (x) + 4-2m = 0 about X has four different real number solutions, then the value range of real number m is?

Given the function f (x) = - | x | + 1, if the equation f ^ 2 (x) + (2m-1) f (x) + 4-2m = 0 about X has four different real number solutions, then the value range of real number m is?

F (x) = - x + 1 = 1-x (1-x) &# 178; + (2m-1) (1-x) + 4-2m = 0 X & # 178; - 2x + 1 + (1-2m) x + 2m-1 + 4-2m = 0 X & # 178; - (2m + 1) x + 4 = 0 [- (2m + 1)] &# 178; - 4 * 4 > 0, (2m + 1) &# 178; - 4 & # 178; > 0, (2m + 1 + 4) (2m + 1-4) > 0, (2m + 5) (2
Given the function f (x) = x (2 + a|x|), and the inequality f (x + a) about X
When x > = 0, f (x) = ax ^ 2 + 2x = a (x + 1 / a) ^ 2-1 / A
When x < 0, G (x) = - ax ^ 2 + 2x = - A (x-1 / a) ^ 2 + 1 / A
When a = 0, a is an empty set, which is rounded off
When a > 0, the opening of quadratic function f (x) is upward, the axis of symmetry x = - 1 / A, f (x) is an increasing function on x > = 0, and a is an empty set
The opening of quadratic function g (x) is downward, the axis of symmetry x = 1 / A, and G (x) is in X
If the odd function f (x) is a decreasing function defined on (- 1,1), find the real number m satisfying f (1-m) + F (1-m & sup2;) < 0
f(1-m)+f(1-m²)
If f (x) is known to be an odd function and f (x) = log2x when x > 0, then the value range of X satisfying the inequality f (x) > 0 is______ .
∵ function f (x) is odd, ∵ f (- x) = - f (x), that is, f (x) = - f (- x), ∵ x < 0, - x > 0, ∵ f (- x) = log2 (- x) = - f (x), that is, f (x) = - log2 (- x), when x = 0, f (0) = 0; ∵ f (x) = log2x, x > 00, x = 0 − log2 (− x), x < 0 When x ＞ 0, the solution of log2x ＞ 0 leads to X ＞ 1. When x ＜ 0, the solution of - log2 (- x) ＞ 0 leads to X ＞ - 1, ■ - 1 ＜ x ＜ 0. To sum up, we can get x ＞ 1 or - 1 ＜ x ＜ 0, so the value range of X is (- 1,0) U (1, + ∞). So the answer is: (- 1,0) U (1, + ∞)
1、 If the odd function f (x) is a decreasing function in the domain (- 1,1), the value range of real number m satisfying f (1-m) + F (M & # 178; - 1) < 0 is obtained
2、 It is known that f (x) is a quadratic function of one variable and satisfies the condition f (x + 1) + F (x-1) = 2x & # 178; - 4x
(1) The analytic expression of function f (x);
(2) F (1 + √ 2)
1, the solution odd function f (x) is a decreasing function in the domain (- 1,1) and satisfies f (1-m) + F (M & # 178; - 1) < 0
That is, f (1-m) + F (M & # 178; - 1) ＜ - f (M & # 178; - 1) = f (1-m & # 178;)
That is - 1 < 1-m < 1-m & # 178; < 1
The solution is 0 < m < 1
Let f (x) = ax ^ 2 + BX + C, from F (x + 1) + F (x-1) = 2x & # 178; - 4x
Know a (x + 1) ^ 2 + B (x + 1) + C + a (x-1) ^ 2 + B (x-1) + C = 2x & # 178; - 4x
2ax²+2bx+2a²+2c=2x²-4x
The solution is a = 1, B = - 2, C = - 1
f(x)=ax^2+bx+c=x^2-x-1
f（1+√2）=（1+√2）²-（1+√2）-1
=3-2√2-√2-2
=1-3√3
Find the range (1) y = x ^ 3-3x ^ 2 + 5, X ∈ [- 2,3] (2) y = 1 / (x + 1) + X, X ∈ [1,3]
Derivation y = 3x ^ 2-6x
When x ∈ [0,2], y < 0, so y decreases monotonically
When x ∈ [- 2,0] or X ∈ [2,3], Y > 0, so y increases monotonically
So the maximum and minimum is the comparison of these points, x = - 2,0,2,3
Two
Homologous derivative
If f (x-1) + F (1-x ^ 2) > 0, then the value range of real number x is?
From F (x-1) + F (1-x ^ 2) > 0 to f (x-1) > - f (1-x ^ 2)
And f (x) is an odd function, - f (1-x ^ 2) = f (x ^ 2-1)
So f (x-1) > F (x ^ 2-1)
And f (x) is a decreasing function defined on [- 1,1]
So x satisfies: - 1 ≤ X-1 ≤ 1, - 1 ≤ x ^ 2-1 ≤ 1, X-1
Find the range of function y = (2 / 3) ^ (- x ^ 2 + 3x-1 / 4)
Let t = - x ^ 2 + 3x-1 / 4,
t=-x^2+3x-1/4=-(x-3/2)^2+2
Because - (x-3 / 2) ^ 2 ≤ 0, so - (x-3 / 2) ^ 2 + 2 ≤ 2
Y = (2 / 3) ^ t is a decreasing function,
So (2 / 3) ^ t ≥ (2 / 3) ^ 2 = 4 / 9,
The function range is [4 / 9, + ∞)
-x^2+3x-1/4=-(x-3/2)²+2
When x = 3 / 2, the maximum denominator is 2
The minimum value of the function is 1 / 3
The range is [1 / 3, + ∞)
If f (m-1) - f (2m-1) > 0, then the solution set of real number m is zero
F (x) is an even function and a decreasing function in [0, + ∝, so it is an increasing function in (- ∝, 0]
f(m-1)-f(2m-1)>0
There are two situations
0>m-1>2m-1
The solution is mm-1 > 0
The solution is m > 1
So the solution set of real number m is (- ∝, 0) ∪ (1, + ∝)
Because the function is even, f (- x) = f (x) = f (| x |)
F (m-1) > F (2m-1) can be reduced to: F (| M-1 |) > F (| 2m-1 |)
Because f (x) decreases monotonically on [0, positive infinity],
So | M-1 | f (2m-1)
1. m>0, m-1 < 2m-1,
F (m-1) > F (2m-1) = > M-1 ≥ 0 and 2m-1 ≥ 0 = > m ≥ 1
2. m 2m-1
F (... Expansion)
Even function f (x) is a decreasing function in [0, + ∞) and an increasing function in (- ∞, 0].
F (m-1) - f (2m-1) > 0 means f (m-1) > F (2m-1)
1. m>0, m-1 < 2m-1,
F (m-1) > F (2m-1) = > M-1 ≥ 0 and 2m-1 ≥ 0 = > m ≥ 1
2. m 2m-1
F (m-1) > F (2m-1) = > M-1 ≤ 0 and 2m-1 ≤ 0 = > m
Find the range of function y = 3x ^ (x Λ 2 + 3x + 1) (x < 0)
The denominator is y + y = 3-y + y = 3
Delta = 9 (Y-1) ^ 2-4y ^ 2 = 5Y ^ 2-18y + 9 > = 0
y>=3 or y
Divide the numerator and denominator by X at the same time, and use the mean inequality at the denominator
Why don't you get to the top?
y=3x/(x^2+3x+1)
It is sorted out as the quadratic equation YX ^ 2 + (3y-3) x + y = 0
Because the domain of the original function is not empty, there is a root for this equation
So the discriminant is (3y-3) ^ 2-4y ^ 2 > = 0
The solution is y = 3
Combined with the actual situation of this function
Because x