It is known that the function f (x) defined on the interval (0, + ∞) satisfies f (x1x2) = f (x1) - f (x2), and when x > 1, f (x) < 0. ① find the value of F (1); ② judge the monotonicity of f (x); ③ if f (3) = - 1, solve the inequality f (| x |) < 2

It is known that the function f (x) defined on the interval (0, + ∞) satisfies f (x1x2) = f (x1) - f (x2), and when x > 1, f (x) < 0. ① find the value of F (1); ② judge the monotonicity of f (x); ③ if f (3) = - 1, solve the inequality f (| x |) < 2

Solution & nbsp; ① Let f (x1x2) = f (x1) - f (x2), let X1 = X2, then f (1) = 0; ② let X1 > x2 > 0, then f (x1) - f (x2) = f (x1x2), because x1x2 > 1, so f (x1x2) < 0, so f (x1) - f (x2) < 0, that is, f (x1) < f (x2), so f (x) is a monotone decreasing function on (0, + ∞); ③ because f (3) = - 1, and f (93) = f (9) - f (3), that is, f (9) = 2F (3) F (| x |) < 2 can be reduced to f (| x |) < f (9), and f (x) is a monotone decreasing function on (0, + ∞), so | x | > 9, the solution is x | 9 or X | 9, so the solution set of F (| x |) < 2 is (- ∞, 9) ∪ (9, + ∞)
It is known that the function f (x) of the domain of definition on the interval (0, + infinity) satisfies f (x1 / x2) = f (x1) - f (x2) and f (x) < 0 when x > 1
If f (3) = - 1, solve the inequality f (the absolute value of x) ＜ - 2
It is known that the function f (x) of the domain of definition on the interval (0, + infinity) satisfies f (x1 / x2) = f (x1) - f (x2) and f (x) < 0 when x > 1
If f (3) = - 1, solve the inequality f (the absolute value of x) ＜ - 2
The function satisfying the given condition is:
f(X) = - log3 X
Inequality: F (x) < - 2
Namely: - log3 x < - 2
The solution is: x > 9
It is known that the function f (x) defined on the interval (0, + ∞) satisfies f (x1x2) = f (x1) - f (x2), and when x > 1, f (x) < 0. ① find the value of F (1); ② judge the monotonicity of f (x); ③ if f (3) = - 1, solve the inequality f (| x |) < 2
Solution & nbsp; ① Let f (x1x2) = f (x1) - f (x2), let X1 = X2, then f (1) = 0; ② let X1 > x2 > 0, then f (x1) - f (x2) = f (x1x2), because x1x2 > 1, so f (x1x2) < 0, so f (x1) - f (x2) < 0, that is, f (x1) < f (x2), so f (x) is a monotone decreasing function on (0, + ∞); ③ because f (3) = - 1, and f (93) = f (9) - f (3), that is, f (9) = 2F (3) F (| x |) < 2 can be reduced to f (| x |) < f (9), and f (x) is a monotone decreasing function on (0, + ∞), so | x | > 9, the solution is x | 9 or X | 9, so the solution set of F (| x |) < 2 is (- ∞, 9) ∪ (9, + ∞)
It is known that the function f (x) defined on the interval (0, + ∞) satisfies f (x1 / x2) = f (x1) - f (x2), and if x > 1, f (x)
Let x = 1, f (1) = f (1) - f (1) = 0
Order 0
f(3)=4
f(3/√3)=f(3)-f(√3)
f(3)=2f(√3)
f(√3)=2
Let X1 > x2
f(x1)-f(x2)=f(x1/x2）
∵x1＞x2 ∴x1/x2＞1
When x > 1, f (x) < 0
∴f(x1)-f(x2）＜0
In the interval (0, + ∞), f (x) is a decreasing function
So a ^ 2 + a-5 > 00
a^2+a-5>√3
a^2+a-5-√3>0
a>{-1+√[1+4(5+√3)]}/2
A
It is known that the function f (x) defined on the interval (0, positive infinity) satisfies f (x1 / x2) = f (x1) - f (x2), and when x is greater than 1, f (x) is less than% d% A. It is known that the function f (x) defined on the interval (0, positive infinity) satisfies f (x1 / x2) = f (x1) - f (x2), and when x is greater than 1, f (x) is less than 0 (1). Find the value of F (1) (2) judge the monotonicity of F (x) (3) if f (3) = - 1, Finding the minimum value of FX on [2,9]
(1) F (1) = f (1) - f (1) = 0. (2) f (x) is a decreasing function at (0, positive infinity). Let x2 & gt; x1, f (x2) - f (x1) = f (x2 / x1). Since x2 / x1 is greater than 1, and X & gt; 1, f (x) & lt; 0, f (x2) - f (x1) is less than 0, so f (x) is a decreasing function. (3) the minimum value is - 2
It is known that the function f (x) defined on (0, positive infinity) satisfies f (x1 / x2) = f (x1) - f (x2) and x > 1, f (x)
(1) Let X1 = X2, then f (1) = 0
(2) Because x ∈ (0, ∞), and f (1) = 0
Let X1 > x2 > 0, because when x > 1, f (x)
The zeros of the function f (x) = 2x & # 178; - 3x + 1 are
A、-1/2,-1 B、1/2,1 C、1/2,-1 D、-1/2,1
If it's a substitution method, it's a single choice question
The solution of 2x & # 178; - 3x + 1 = 0 is (2x-1) (x-1) = 0, and x = 1 / 2,1
B is the answer
1 / 2, 2 * 1 / 4-3 / 2 + 1 = 0. What's the problem?
1, 2-3 + 1 = 0
How many zeros does the function f (x) = (1 / 2) ^ x + 3x ^ 2-2 have?
How to judge?
The zero point of F (x) = (1 / 2) ^ x + 3x ^ 2-2 is that,
（1/2)^x+3x^2-2=0
You change it into:
3x^2-2=（1/2)^x
You can draw the general picture of F (x) = 3x ^ 2-2 and f (x) = (1 / 2) ^ X and see how many intersections there are. Because you can't solve this equation, you have to use the picture
Finding the zero point of function y = (x-1) (x ^ 2-3x + 1)
y=(x-1)(x²-3x+1)=0
x-1=0,x²-3x+1=0
So x = 1, x = (3 - √ 5) / 2, x = (3 + √ 5) / 2
y=(x-1)(x^2-3x+1)=0
have to
Zero is zero
X = 1 or x = (3 ± √ 5) / 2
What grade is this?
The function f (x) = ln (1 / 20) ^ x-lgx is known if the real number x0 is the zero point of function f (x) and 0
X2 should be x0
f(x)=xln(1/20)-lgx
=-(lg20)x-lgx
-(LG20) x and - lgx are both decreasing functions
So f (x) decreases
x1f(x0)=0
Choose a