It is known that f (x) is an odd function defined on the interval [- 1,1], and f (1) = 1. If m, n Euro [- 1,1], M + n is not equal to zero, then [f (m) + F (n)] / (M + n) > 0 Prove that f (x) is an increasing function on [- 1,1] Solving inequality (x + 1 / 2) < f (1 / x-1)

Because f (x) is an odd function on [- 1,1], we have f (- x) = f (x), let x 10, so
[f (x1) + F (- x2)] / x1-x2 > 0, and because x1-x2

Given that f (x) is an odd function defined on [- 1,1], and f (1) = 1, if m, n ∈ [- 1,1], M + n is not equal to 0, then
[f (m) + F (n)] / M + n > 0, if f (x)
T2 = T & #, wrong number~

[f (m) + F (n)] / M + n > 0, so: (f (m) + F (- n)) / (m-n) > 0f (- n) = - f (n) (f (m) - f (n)) / (m-n) > 0f (m) - f (n) has the same sign as M-N, that is, when m > N, f (m) > F (n) so the function is an increasing function. In the interval [- 1,1], the maximum value of F (x) = f (1) = 1, if f (x) = 1 / 2, there is no solution at the same time

It is known that the domain of definition of function FX is R. for any real number m, n satisfies the following conditions: F 1 / 2 = 2, and f (M + n) = f (m) + F (n)
(1) Finding the value of F minus 1 / 2 (2) proving that FX is a monotone decreasing function on the domain R

In order to use the law of kibiser, f (M + n) = f (m) + F (n) is reduced to f * m + f * n = V * m + V * n
Then confirm that the value 2 in 1 / 2 is a positive function and a negative function
(for bonus points)

It is known that f (x) is an odd function defined on the interval [- 1,1], and f (1) = 1. If m, n Euro [- 1,1], M + n is not equal to zero, then [f (m) + F (n)] / (M + n) > 0 Prove that f (x) is an increasing function on [- 1,1] Solving inequality (x + 1 / 2) < f (1 / x-1)