Let f (x) be a decreasing function defined on (- 2,2) and satisfy the following conditions: F (- x) = - f (x), and f (m-1) + F (2m-1) > 0, then the value range of real number m is obtained

Let f (x) be a decreasing function defined on (- 2,2) and satisfy the following conditions: F (- x) = - f (x), and f (m-1) + F (2m-1) > 0, then the value range of real number m is obtained

The inequality f (m-1) + F (2m-1) > 0, that is, f (m-1) > F (2m-1), ∵ f (- x) = - f (x), we can get - f (2m-1) = f (- 2m + 1) ∵ the original inequality is transformed into f (m-1) > F (- 2m + 1) and ∵ f (x) is a decreasing function defined on (- 2,2), ∵ - 2 < M-1 < - 2m + 1 < 2, the solution is - 12
Find the range of function y = 3 + √ (2-3x)
RT
2-3x≥0
therefore
The range is
y≥3+0=3
That is [3, + ∞)
Given the function f (x) = {2 ^ X-1, x > 0 - x ^ 2-2x, X ≤ 0, if the function g (x) = f (x) - M has three zeros, then the value range of real number m
I calculate it is [0,1], but the answer is (0,1). Why is m not equal to 0
Is f (x) = {2x-1, x > 0 - x ^ 2-2x, X ≤ 0
X & lt; = 0, - x ^ 2-2x = - (x + 1) ^ 2 + 1 & lt; = 1, G (x) & lt; = 1-m, X & gt; 0, & nbsp; 2 ^ X-1 & gt; 0, G (x) & gt; = - m, when G (x) = 0, f (x) can satisfy three X, one is greater than 0, two is less than 0; that is to say, the coincidence interval of F (x) piecewise function value is (0,1) why can not be 0 for the landlord's question, if M = 0, then the solution of 2 ^ X-1 = 0 follows x = 0, which does not satisfy the value interval of F (x)
The range of function f (x) = radical x + radical 6-3x
How do you calculate it,
f(x)=√x+√(6-3x)
The domain is x > = 0,6-3x > = 0, that is, 0=
If x ∝ (1 / 3,3), then | logax|
(1 / 3,1) U (3, positive infinity)
Find the range of function f (x) = - 2x ^ 2 + 3x + 1 x ∈ [- 3 / 2,2]
The function f (x) is a quadratic function, its axis of symmetry is x = 3 / 4, and its opening is downward
When x = 3 / 4, the maximum value of F (x) is 17 / 8
-The distance from 3 / 2 to the axis of symmetry is greater than the distance from 2 to the axis of symmetry
So when x = - 3 / 2, f (x) takes the minimum value of - 8
When x ∈ [- 3 / 2,2], the range of F (x) is [- 8,17 / 8]
Calculate the axis of symmetry to see if it is in the given domain~~~
If you don't bring in two numbers, OK
If it must be the maximum or minimum at the vertex, (the opening of the parabola is downward, so it is the maximum)
Then take those two points and find the maximum or minimum
Formula drawing
If the function y = logax always has | y | 1 on X ∈ [2, + ∞), then the value range of a is ()
A. 0 < a < 12 or 1 < a < 2B. 12 < a < 1 or 1 < a < 2C. 1 < a < 2D. 0 < a < 12 or a > 2
∵ function y = logax always has y | y | 1 on X ∈ [2, + ∞); (1) when 0 ＜ a ＜ 1, function y = logax always has y ＜ - 1 on X ∈ [2, + ∞); (2) when a ＞ 1, function y = logax always has y ＞ 1 on X ∈ [2, + ∞); (2) when a ＞ 1, function y = logax always has y ＞ 1 on X ∈ [2, + ∞); (2) from (1) and (2) we can get 12 ＜ a ＜ 1 or 1 ＜ a ＜ 2, so we should choose B
It is known that f (x) is an odd function. When x ∈ (0,1), f (x) = lg11 + X, then when x ∈ (- 1,0), the expression of F (x) is______
When x ∈ (- 1,0), - x ∈ (0,1) ∵ f (- x) = lg11 − x = - LG (1-x). ∵ f (x) is an odd function, f (- x) = - f (x), that is - f (x) = - LG (1-x). When x ∈ (- 1,0), f (x) = LG (1-x), so the answer is: F (x) = LG (1-x)
Given the function f (x) = logax-2x ^ 2, if x ∈ (0,1 / 2), f (x) > 0, find the value range of real number a
Because x > 0 a > 0 a is not 1
So when 0
∵x－2x²＝﹣2(x－1/4)²＋1/8
∵x∈(0,1/2) ∴x－2x²∈(0，1/8)
∵f(x)＞0 ∴0＜a＜1
If the function y = f (x) is odd and f (x) = LG (x + 2) when x > 0, then f (x) =?
X0, so
f(-x)=lg(-x+2)
Because f (- x) = - f (x)
So f (x) = - LG (- x + 2)
therefore
f(x)=lg(x+2)(x>0).or.-lg(2-x)(x
f(x)=lg(x+2)x>0
lg(-x+2)0>x
f(x)=lg(x+2),
Y = f (x) is an odd function
f(-x)=lg(-x+2)