Given the function FX = (PX ^ 2 + 2) / (q-3x), find the monotone interval of the function Please look at the title, The analytic expression is FX = (2x ^ 2 + 2) / (- 3x)

Given the function FX = (PX ^ 2 + 2) / (q-3x), find the monotone interval of the function Please look at the title, The analytic expression is FX = (2x ^ 2 + 2) / (- 3x)

Function FX = x + 1 / x, judge the monotonicity of FX in (0,1) and prove it
Note: FX = x + (1 / x) derivative is unknown
Set 0
Given the function f (x) = x3-3x 1, find the value of F '(2), find the monotone interval of function FX, and find the extremum and maximum value of function FX in - 22
Given the function f (x) = x3-3x 1, find the value 2 of F '(2), find the monotone interval 3 of function FX, and find the extremum and maximum value of function FX in [- 2,2]
F & # 39; (x) = 3x ^ 2-3, so f & # 39; (2) = 9; F & # 39; (x) & gt; = 0, we get X & gt; = 1 or X & lt; = - 1, so the monotone increasing interval of FX is (negative infinity, - 1), [1, positive infinity), monotone decreasing interval is [- 1,1]. F (- 2) = - 2, f (2) = 2, and because FX is in [- 2, - 1], [1,2] monotone increasing, in [- 1,1] monotone
Given the function f (x) = loga (1 + X / 1-x), (1) find the domain of definition of F (x); (2) judge the parity of F (x) and prove it;
(3) Find the value range of x such that f (x) > 0
(1) If the domain is (1 + x) / (1-x) > 0 and 1-x is not equal to 0, the solution is - 11 and the solution is 0
The monotone decreasing interval of the function y = xcube-3x is
Solution;
y‘=3x²-3
When y '
Just take the derivative.
y‘=3x²-3
Let y '= 0, get x = - 1 and 1
When x ∈ (- 1,1), y '< 0, y decreases monotonically on (- 1,1).
Given the function f (x) = x + 1 / x, judge the monotonicity of F (x) in the interval [1, + ∞], and explain the reason
Let 1 "X1"
The zero point of function f (x) = 2x + 3x is in an interval ()
A. （-2，-1）B. （-1，0）C. （0，1）D. （1，2）
From F (− 1) = 12 − 3 ＜ 0, f (0) = 1 ＞ 0, and the zero point theorem, we know that the zero point of F (x) is in the interval (- 1, 0), so we choose B
It is known that the definition field of function f (x) is r, and for any real number x1, X2, there is always f (x1 + x2) + F (x1-x2) = 2F (x1) f (x2). It is proved that f (x) is even function
Let X1 = - 1, X2 = 1
Then f (- 1) = f (1) = f (- 1) so
f（1）=0
Let X1 = - 1, X2 = - 1
Then f (1) = 2F (- 1)
So f (- 1) = 0
Let X1 = x, X2 = - 1, X belong to its domain
Then f (- x) = f (x) + F (- 1) = f (x)
So f (x) is even function
Given that the function FX is an odd function defined at (- 2,5), find the increasing interval of function f (6-3x)
Is the function f (x) a decreasing function defined at (- 2,5)?
Composite function problem
First find the domain - 2
Let f (x) define the field (0, + ∞), and f (4) = 1. For any positive real number x1x2, f (x1x2) = f (x1) + F (x2), and if x1 ≠ X2, f (x1x2) = f (x1) + F (x2)
Let f (x2) - f (x1) / x2-x1 > 0 (1) judge the monotonicity of function f (x) on (0, + ∞) (2) find the value of F (1) (3) if f (x + 6) + F (x) > 2, find the value range of X
(1) Because f (x2) - f (x1) / x2-x1 > 0, it can be divided into two categories: a. the numerator denominator is greater than 0. B. the numerator denominator is less than 0
(2) Let X1 = x2 = 1, substitute f (x1x2) = f (x1) + F (x2), f (1) = 0
(3) According to f (x1x2) = f (x1) + F (x2), then f (x + 6) + F (x) = f [(x + 6) * x], and because f (4) = 1, so f (4 * 4) = f (4) + F (4) = 4, that is, f (16) = 2. So the inequality is transformed into f [(x + 6) * x] > F (16). Because the function is an increasing function, so [(x + 6) * x > 16, solve this one variable quadratic inequality, and pay attention to the condition of x > 0. The final result is x > 2