Given the function f (x) = Log1 / 2 (x ^ 2-mx-m) (1) if M = 1, find the domain of definition of function f (x) (2) Given function f (x) = Log1 / 2 (x ^ 2-mx-m) (1) If M = 1, find the domain of F (x) (2) If the value range of function f (x) is r, find the value range of real number M (3) If the function f (x) is an increasing function in the interval (- ∞, 1-radical 3), the range of the real number m is obtained The process of the third question

Given the function f (x) = Log1 / 2 (x ^ 2-mx-m) (1) if M = 1, find the domain of definition of function f (x) (2) Given function f (x) = Log1 / 2 (x ^ 2-mx-m) (1) If M = 1, find the domain of F (x) (2) If the value range of function f (x) is r, find the value range of real number M (3) If the function f (x) is an increasing function in the interval (- ∞, 1-radical 3), the range of the real number m is obtained The process of the third question

x^2-mx-m=x^2-x-1>0
x> (1 + radical 5) / 2, x = 0
m> = 0, M = = 1-radical 3
m> = 2 - (2 * radical 3)

If f (x) is an odd function with a period of 4, and f (- 1) = a, (a ≠ 0), then the value of F (5) is equal to?

f(3)=f(-1+4)=a
f(-3)=-a
f(5)=f(-3+4+4)=-a

If f (x) is an odd function with a period of 4 and f (- 1) = a (a is not equal to 0), then f (5) has the same value

Because f (x) is an odd function
So f (- 1) = - f (1)
So f (1) = - A
And because f (x) is a periodic function with period 4, f (5)) = f (1 + 4) = f (1)
So f (5) = - A

How to prove the derivative of function f (x) = a ^ x?
Can we give a detailed proof?
(a^(x+h)-a^x)/h = a^x * (a^h-1)/h -> a^x * ln(a)

This has a little skill. You can refer to any "mathematical analysis" textbook. The method is almost the same
My textbook does this:
In this paper, we first prove that the derivative of G (x) = x ^ A is ax ^ (A-1). In fact, if x ≠ 0, then there is
(g(x+h)-g(h))/h = x^(a-1)*((1+h/x)^a-1)/(h/x)
For fixed x ≠ 0, because when h - > 0, H / X - > 0
g'(x) = ax^(a-1)
Secondly, for f (x) = a ^ x, when h - > 0
(a^(x+h)-a^x)/h = a^x * (a^h-1)/h -> a^x * ln(a)
So the conclusion is drawn
Note: the previous step used the limit
((1+x)^a-1)/x -> a
(when X - > 0)
You can try the important limit (1 + x) ^ (1 / x) - > e to do this

If the function f (x) is monotone in the interval [a, b], and f (a) * f (b)

Because f (x) is a monotone function on the interval [a, b], and f (a) * f (b)

The definition field of function f (x) is r, and f (x) = f (6-x). If the equation f (x) = 0 has four roots, then the sum of the four roots is

f(x)=0,
If f (x1) = 0, then f (6-x1) = 0,
X1,6-x1 are the two roots of the equation f (x) = 0;
If the equation f (x) = 0 has four roots, then:
If f (x2) = 0, then f (6-x2) = 0,
X 2,6-x 2 are also two roots of the equation f (x) = 0
So X1 + (6-x1) + x2 + (6-x2) = 12

The definition domain of function f (x) is r, and f (x) = f (12-x). The equation f (x) = 0 has n real roots, and the sum of these n real roots
So what is n

Obviously, if a is a root of the equation, then 12-A must also be a root of the equation, that is, the root of the equation is symmetrically distributed with respect to x = 6, then the sum of n roots is 6N
6n=1992
n=332

Given that y = f (x) is an odd function defined on R, when x > 0, f (x) = X-2, then the solution set of inequality f (x) < 0 is ()
A. （0，+∞）B. （-2，2）C. （-∞，-2）∪（2，+∞）D. （-∞，-2）∪（0，2）

① When x ＞ 0, f (x) = X-2, then when x ＞ 2, f (x) ＞ 0, when 0 ＜ x ＜ 2, f (x) ＜ 0; and ∪ y = f (x) is an odd function defined on R, and the solution set of the inequality f (x) ＜ 0 is (- ∞, - 2) ∪ (0, 2)

Given that the function y = f (x) is a decreasing function defined on (- 8, + positive infinity), and f (0) = 0, find the inequality f (x ^ 2-4x-5) > 0

Y = f (x) is a decreasing function defined on (- 8, + positive infinity), and f (0) = 0, so the solution of y = f (x) > 0 is 0 > x > - 8, so the solution of y = f (x ^ 2-4x-5) > 0 is 0 > x ^ 2-4x-5 > - 8, that is, 0 > (X-2) ^ 2-9 > - 89 > (X-2) ^ 2 > 13 > | X-2 | 1 3 > X-2 > 1 or - 1 > X-2 > - 3, so the solution is: 5 > x > 3 or 1 > X

It is known that f (x) is an odd function over R, and for any real number x, f (x + 3) + F (x) = 0. If x belongs to [- 3, - 2], f (x) = 2x, find f (1 / 2)

Let x = - 1 / 2, then f (- 1 / 2 + 3) + F (- 1 / 2) = 0
That is, f (- 1 / 2) = - f (5 / 2)
∵ f (x) is an odd function over R
∴f(-1/2)=-f(1/2),-f(5/2)=f(-5/2)
∴-f(1/2)=f(-5/2) ∴f(1/2)=-f(-5/2)
If ∵ x ∈ [- 3, - 2], f (x) = 2x
∴f(-5/2)=2(-5/2)=-5
∴f(1/2)=-f(-5/2)=5