It is known that f (x) is an even function, which is a subtractive function on [0, + ∞). If f (lgx) > F (1), the value range of X is

It is known that f (x) is an even function, which is a subtractive function on [0, + ∞). If f (lgx) > F (1), the value range of X is

Even function
f(x)=f(-x)
That is, f (x) = f (|x|)
So f|lgx|) > F (1)
X > 0 decrement
So lgx|

It is known that f (x) is an even function and it is a subtractive function on [0, + 00). If f (lgx) > F (1), what is the value range of X

Because f (x) is an even function, f (- x) = f (x), f (1) = f (- 1)
F (x) is a decreasing function on [0, + ∞), then f (x) is an increasing function on (0, -∞),
When lgx > = 0, if f (lgx) > F (1), then 0

It is known that f (x) is an even function, which is a subtractive function on [0, + ∞). If f (lgx) > F (1), the value range of real number x is () A. （1 10，1） B. （0，1 10）∪（1，+∞） C. （1 10，10） D. （0，1）∪（10，+∞）

∵ f (x) is an even function, which is a subtractive function on [0, + ∞),
‡ f (x) monotonically increases on (- ∞, 0),
From F (lgx) > F (1), f (1) = f (- 1)
Get: - 1 ＜ lgx ＜ 1,
∴1
10＜x＜10，
So the answer is C

. f (x) is an odd function on R and an increasing function on the interval (negative infinity to 0), f (2) = 0, then the solution of inequality x * f (x) > 0 F (x) is an odd function on R and an increasing function on the interval (negative infinity to 0). If f (2) = 0, the solution of inequality x * f (x) > 0 If the primary function f (x) is a subtractive function and f [f (x)] = 4x-1, its analytical formula is f (x)=

Let the function f (x) = x * f (x),
Because f (x) is an increasing function on the interval (negative infinity to 0), it is also an increasing function on [0, + ∞)
Because f (2) = 0, it can be seen that f (- 2) = 0
When x0, and x > 0
So x * f (x) > 0
When x

If f (x) is an increasing function defined on (0, + infinity), then the solution set of inequality f (x) > F [8 (X-2)] is?

Because it's an increasing function
Because f (x) > F [8 (X-2)]
So x > 8 (X-2)
So x ＜ 16 / 7
Because x (0, infinity)
therefore
0〈x〈16/7

Given that the odd function f (x) satisfies f (- 1) = 0 and is an increasing function from 0 to positive infinity, the solution set of inequality XF (x) < 0

Because f (x) is an increasing function on (0, + infinity),
So, f (x) is also an increasing function on (- infinity, 0),
And f (x) is an odd function, f (- 1) = 0, so f (1) = 0,
And when x < - 1, f (x) < 0- 10； f(0)=0； At 01, f (x) > 0,
X * f (x) < 0, then x is different from F (x),
Therefore, the solution set is: (- 1,0) U (0,1)