The definition field of F (x) is known Given the domain [- 2,2], for any x belonging to [- 2,2], there is f (- x) = - f (x), and f (2) = 2. For any m, n belonging to [- 2,2], M + n is not equal to 0, there is f (m) + F (n) / M + n > 0 (1) It is proved that f (x) is an increasing function on [- 2,2] by definition; (2) Solving inequality f (x + 1 / 2)

The definition field of F (x) is known Given the domain [- 2,2], for any x belonging to [- 2,2], there is f (- x) = - f (x), and f (2) = 2. For any m, n belonging to [- 2,2], M + n is not equal to 0, there is f (m) + F (n) / M + n > 0 (1) It is proved that f (x) is an increasing function on [- 2,2] by definition; (2) Solving inequality f (x + 1 / 2)

(1)
f(x+△x)-f(x)
=f(x+△x)+f(-x)
[f(x+△x)+f(-x)]/[(x+△x)+(-x)]>0
f(x+△x)+f(-x)>0
f(x+△x)-f(x)>0
F (x) is an increasing function on [- 2,2]
(2)
f[(x+1)/2]

If the domain of function y = f (x) is [- 2,4], then the domain of function g (x) = f (x) + F (- 1) is?

The domain of the function g (x) = f (x) + F (- 1) is [- 2,4]
Because f (- 1) is a constant, the domain of G (x) is the domain of F (x)

If the n power of 3 × 9 × 27 × 3 is equal to the 8 power of 3, what is n equal to

Original formula = 3 ^ (1 + 2 + 3 + n) = 3 ^ 8
So:
6+n=8
n=8-6=2
So n = 2
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One half plus one half of the second power plus one half of the third power, and so on to the nth power. The result is expressed by the formula containing n

This is an equal ratio sequence, and the common ratio is half
We can directly set the formula to get one of the nth power of 1-2, n from 1 to infinity
Don't you understand
One minus two to the nth power, n goes from one to infinity
That's the answer