Let f (x) = e ^ x + SiNx. G (x) = 1 / 3x. If x 1 and x 2 belong to 0 to positive infinity, if f (x 1) = g (x 2), then x 1-x 2 is the minimum

Let f (x) = e ^ x + SiNx. G (x) = 1 / 3x. If x 1 and x 2 belong to 0 to positive infinity, if f (x 1) = g (x 2), then x 1-x 2 is the minimum

x> The derivative of function f (x) is e ^ x + cosx > 0 (e ^ x > 1, cosx ≥ - 1), and the function f (x) increases monotonically. In this interval, G (x) is a monotone decreasing function. The approximate image of two functions can be drawn based on this. To find the minimum value of x1-x2, we need to make X1 minimum and X2 maximum

It is known that the function f (x) satisfies (x1-x2) [f (x1) - f (x2)] > 0 on [0, + ∞), and f (2x-1)

(x1-x2) [f (x1) - f (x2)] > 0, that is, when X1 > X2, f (x1) > F (x2), so the function is an increasing function in the domain x > = 0
So f (2x-1) - 1
Domain requirements:
2x-1>=0 -->x>=1/2
3x>=0 -->x>=0
The conclusion is: x > = 1 / 2

The domain of F (x) is symmetric with respect to the origin, and f (x1-x2) = [f (x1) f (x2) + 1] / F (x2) - f (x1), there is a normal number a such that f (a) = 1
Prove: 1, f (x) is an odd function
2, f (x) is a periodic function and the period is 4a

1. F (x1-x2) = [f (x1) f (x2) + 1] / [f (x2) - f (x1)] f (x2-x1) = [f (x1) f (x2) + 1] / [f (x2) - f (x1)]. Obviously, f (x1-x2) + F (x2-x1) = 0 is an odd function because the domain of definition is symmetric with respect to the origin. 2. Let X1 = x + A, then f (a) = [f (x) f (x + a) + 1] / [f (x) - F (x + a)] = 1 to simplify f (x +

Let the domain of F (x) be symmetric about the origin, and for any x1 ≠ x2 in the domain, f (x1-x2) = [1 + F (x1) + F (x2)] / [f (x2) - f (x1)]
Prove that f (x) is an odd function

f(x2-x1)=[1+f(x1)+f(x2)]/[f(x1)-f(x2)]=-[1+f(x1)+f(x2)]/[f(x2)-f(x1)]=-f(x1-x2)
It is an odd function