According to the definition of function monotonicity, it is proved that the function F & nbsp; (x) = - X3 + 1 is a decreasing function on (- ∞, + ∞)

According to the definition of function monotonicity, it is proved that the function F & nbsp; (x) = - X3 + 1 is a decreasing function on (- ∞, + ∞)

Proof: proof 1: take x1, X2 in (- ∞, + ∞) and X1 < X2, then f (x2) - f (x1) = x13-x23 = (x1-x2) (X12 + x1x2 + X22) ∵ X1 < X2, ∥ x1-x2 < 0. When x1x2 < 0, there is X12 + x1x2 + X22 = (x1 + x2) 2-x1x2 > 0; when x1x2 ≥ 0, there is X12 + x1x2 + X22 > 0; ∥ f (x2) - f (x1) = (x1-x2) (X12 + x1x2 + X22) < 0 Let f (x) = -x3 + 1 be a decreasing function on (- ∞, + ∞) at (- ∞, + ∞) and (x 1, x 2, and x 1 ＜ X2, then f (x2) - f (x2) - f (x1) - f (x1) = f (x2) - f (x2) - f (x2) - f (x1) = f (x2) x (x (x (x) = (x1-x2) (x1-x2) (x1-x2) (x1-x2) (x1-x2) (x1-x2) (x1-x2-x2) (x1-x2-x2-2 + X-2) (x (x-12-2-2-x (x (x (x (x (x (x (x (x (x (x (x (x) = (x (x (x (x (x (x (x (x (x (x (x (x (x (x (x (x (x (x (x))))))))))) (x (x (x (x (x (x (x (x (x (x (2 it's not easy Therefore, the function f (x) = - X3 + 1 is a decreasing function on (- ∞, + ∞)
When y = f (x) is an odd function and x0, f (x)=
RT
Y = f (x) is an odd function,
When X0
-x
Given the function f (x) = x + 4 / x, the monotonicity of the function on (2, positive infinity) is judged and proved by the definition of monotonicity
Monotonic increase
Using the definition of monotonicity,
Let X1 > x2 > 2, so x1-x2 > 0,
F(X1)-F(X2)=X1+4/X1-(X2+4/X2)
=X1-X2+4/X1-4/X2
=(X1-X2)+4(X2-X1)/X1·x2
=(x1-x2)·(1-4/x1·x2)
=(x1-x2)·(x1·x2-4)/x1·x2 (*)
From the above formula: x1-x2 > 0, x1 · x2-4 > 0
(*) is greater than zero, so f (x1) - f (x2) > 0
So it increases monotonically from two to positive infinity
Given that the function y = f (x) is odd, if x > 0, f (x) = LG (x + 1), find f (x)
The domain of definition is r
Let x < 0, then - x > 0, f (- x) = LG (- x + 1) = - f (x)
When x < o, f (x) = - LG (- x + 1)
Oh, don't forget x = 0
According to the definition of function monotonicity, it is proved that the function F & nbsp; (x) = - X3 + 1 is a decreasing function on (- ∞, + ∞)
Proof: proof 1: take x1, X2 in (- ∞, + ∞) and X1 < X2, then f (x2) - f (x1) = x13-x23 = (x1-x2) (X12 + x1x2 + X22) ∵ X1 < X2, ∥ x1-x2 < 0. When x1x2 < 0, there is X12 + x1x2 + X22 = (x1 + x2) 2-x1x2 > 0; when x1x2 ≥ 0, there is X12 + x1x2 + X22 > 0; ∥ f (x2) - f (x1) = (x1-x2) (X12 + x1x2 + X22) < 0 Let f (x) = -x3 + 1 be a decreasing function on (- ∞, + ∞) at (- ∞, + ∞) and (x 1, x 2, and x 1 ＜ X2, then f (x2) - f (x2) - f (x1) - f (x1) = f (x2) - f (x2) - f (x2) - f (x1) = f (x2) x (x (x (x) = (x1-x2) (x1-x2) (x1-x2) (x1-x2) (x1-x2) (x1-x2) (x1-x2-x2) (x1-x2-x2-2 + X-2) (x (x-12-2-2-x (x (x (x (x (x (x (x (x (x (x (x (x) = (x (x (x (x (x (x (x (x (x (x (x (x (x (x (x (x (x (x (x))))))))))) (x (x (x (x (x (x (x (x (x (x (2 it's not easy Therefore, the function f (x) = - X3 + 1 is a decreasing function on (- ∞, + ∞)
On the function f (x) = LG [(x ^ 2 + 1) / | x |] (x is not equal to 0, X belongs to R)
A. The image of function y = f (x) is symmetric about y axis
B. On the interval (negative infinity, 0), the function f (x) is a decreasing function
C. The minimum value of function f (x) is LG2
D. In the interval (1, positive infinity), the function f (x) is an increasing function
The correct proposition is? Please analyze, not just the answer
Questioner: study hard 520 - Intern level
It is transformed into f (x) = LG [| x | + 1 / | x |]
It can be seen that the real number is a typical Nike function
It's just that x is added to the absolute value
Then it's easy
Because the real number is | x | + 1 / | X|
So f (- x) = f (x)
So about Y-axis symmetry
① Yes
Let's not worry about LG itself
So look at the real number
When x ＞ 0, the real number is x + 1 / X. when x ＞ 0, it's a Nike function. Here, you can draw a picture yourself
Because f (- x) = f (x), it is even
When x is less than 0, it is symmetrical about the Y axis
We can see that when x ∈ (0,1) ∪ (1, + infinity) increases
X ∈ (1,0) ∪ (0,1) minus
So it's wrong
Because x ＞ 0 x + 1 / X ≥ 2, root sign x + 1 / x = 2
Even function again
So when x = 1, take the minimum LG2
So, right
It is known that the function f (x) is an increasing function on R and G (x) is a decreasing function on R. this paper proves that the function f (x) = f (x) - G (x) is an increasing function on R
Let x 1, x 2 ∈ R, and x 1 ＜ x 2, then f (x) is an increasing function on R and G (x) is a decreasing function on R, where f (x 1) ＜ f (x 2), G (x 1) ＞ g (x 2), f (x 1) - f (x 2) = [f (x 1) - G (x 2)] = [f (x 1) - f (x 2)] - [g (x 1) - G (x 2)] < 0, f (x 1) ＜ f (x 2) is an increasing function on R
The function f (x) = x ^ 2 + (a + 1) x + LG | a + 2 | (a ∈ R, and a is not equal to - 2)
(2) If f (x) and G (x) are decreasing functions in the interval (- ∞), (a + 1) ^ 2), the value range of a is obtained
(3) Under the condition of (2), compare the size of F (1) and 1 / 6
I can only see the second and third questions (2) because LG | a + 2 | is a constant, the derivation of F (x) leads to f '(x) = 2x + A + 1. Because f (x) is a decreasing function in the interval (- ∞, (a + 1) ^ 2), Let f' (x) = 2x + A + 10 ≤ 0, then x ≤ - 1 / 2 (a + 1), so (a + 1) ^ 2 ≤ - 1 / 2 (a + 1) leads to 2A & # 178; + 5A + 3 ≤ 0, that is (2
It is proved that the function f (x) = x + 1 / X is a decreasing function on (0,1),
Let x1, X2 be on (0,1) and x1
F (x) = (x + 1) / x, let X1 and X2 belong to (0,1) and X11
So f (x) = x + 1 / X is a decreasing function on (0,1)
F (x) = x + 1 / x = 1 + 1 / X on (0,1), 1 / X is a part of hyperbola, x increases gradually from 0 to 1, and the value of 1 / X gradually approaches 1 from positive infinity, that is, a decreasing function, then 1 + 1 / x is equivalent to moving 1 / x up the Y axis, and its value decreases from positive infinity to close to 2 at (0,1), so it is a decreasing function.
Let X1 > X2, and belong to (0,1), f (x1) - f (x2) = X1 + 1 / x1-x2-1 / x2 = X1 ^ 2x2 / x1x2 + x2 / x1x2-x1x2 ^ 2-x1 / x1x2 = X1 ^ 2x2 + x2-x1x2 ^ 2-x1 / x1x2 = x1x2 (x1-x2) + (x2-x1) / x1x2 = (x1x2-1) (x2-x1) / x1x2. Since the definition field is (0,1), x1x2 > 0, (x2-x1) > 0, (x1x2-1) is less than 0 (here, because X1... Is expanded)
Let X1 > x2 and belong to (0,1), f (x1) - f (x2) = X1 + 1 / x1-x2-1 / x2 = X1 ^ 2x2 / x1x2 + x2 / x1x2-x1x2 ^ 2-x1 / x1x2 = X1 ^ 2x2 + x2-x1x2 ^ 2-x1 / x1x2 = x1x2 (x1-x2) + (x2-x1) / x1x2 = (x1x2-1) (x2-x1) / x1x2. Since the definition field is (0,1), x1x2 > 0, (x2-x1) > 0, (x1x2-1) is less than 0 (here, because x1x2 are all less than 1, the product must be less than 1, and then 1 is less than 0), so the whole formula is less than 0, then f (x1) - f (x2) is less than 0, so the function f (x) = x + 1 / X is a decreasing function on (0,1). Put it away
Given the function f (x) = x2 + (a + 1) x + LG | a + 2 | (a ∈ R, and a ≠ - 2) (I) if f (x) can be expressed as the sum of an odd function g (x) and an even function H (x), find the analytic expressions of G (x) and H (x); (II) proposition p: the function f (x) is an increasing function in the interval [(a + 1) 2, + ∞); proposition q: the function g (x) is a decreasing function. If proposition p and Q have and only one is an increasing function Under the condition of (II), compare the size of F (2) and 3-lg2
(1) ∵ f (x) = g (x) + H (x), G (- x) = - G (x), H (- x) = H (x) ∵ f (- x) = - G (x) + H (x) = x2 + (a + 1) x + LG | a + 2 | - G (x) + H (x) = X2 - (a + 1) x + LG | a + 2 |, G (x) = x (a + 1) x, H (x) = x2 + LG | a + 2 | (II) ∵ function f (x) =