If f (x) is a function x (R) = 0, and f (x / R) = 0, then f (f) is a function x (R) = 0, F (101 / 17), f (106 / 15),

If f (x) is a function x (R) = 0, and f (x / R) = 0, then f (f) is a function x (R) = 0, F (101 / 17), f (106 / 15),

F (98 / 19) = f (22 / 19), f (101 / 17) = f (33 / 17), f (106 / 15) = f (16 / 15). We can see the size relationship from the function image. F (101 / 17) > F (98 / 19) > F (106 / 15)

When x > 0, f (x) = x2 power - 2x, find f (x)
Even function defined on r y = f (x), when x > 0, f (x) = x2 power + 2x, find f (x)
2. Prove that f (x) = x + 1 / X is a monotone increasing function on (1, + infinity)

1 (1) odd function satisfies f (x) = - f (- x)
-f(-x)=-（(-x)²-2（-x））=-x²-2x
So when f (x) = x & sup2; - 2x x > 0, f (x) = - X & sup2; - 2x x0, f (x) = x & sup2; - 2x x1, f (x) > 0
So f (x) is a monotone increasing function on (1, + infinity)

Given the function g (x) = f (x) - 1 / F (x), where log2f (x) = 2x, X belongs to R, then what function is g (x)? The answer is even function and increasing function
There are steps to solve the problem

Log2f (x) = 2x f (x) = 2 ^ 2x = 4 ^ XG (x) = f (x) - 1 / F (x) = 4 ^ X - (1 / 4) ^ x, so g (- x) = 4 ^ - X - (1 / 4) ^ - x = (1 / 4) ^ x-4 ^ x = - G (x), so it is an odd function g '(x) = (4 ^ x) ln4 - [(1 / 4) ^ x] ln (1 / 4) = [4 ^ x + (1 / 4) ^ x] ln4 > 0, so it is an increasing function