If the function f (x) defined on R satisfies f (M + n) = f (m) + 2 [f (n)] ^ 2, where m and N belong to R, f (1) ≠ 0, then f (100) =?

If the function f (x) defined on R satisfies f (M + n) = f (m) + 2 [f (n)] ^ 2, where m and N belong to R, f (1) ≠ 0, then f (100) =?

f(2) = f(1+1) = f(1) + 2f(1)^2
f(3) = f(2+1) = f(2) + 2f(1)^2 = f(1) + 4f(1)^2
f(3) = f(1+2) = f(1) + 2f(2)^2 = 8f(1)^4 + 8f(1)^3 + 2f(1)^2 + f(1)
8f(1)^4 + 8f(1)^3 + 2f(1)^2 + f(1) = f(1) + 4f(1)^2
4f (1) ^ 2 + 4f (1) - 1 = 0, the solution is f (1) = - 1 ±√ 2
2f(1)^2 = 3±2√2
So f (100) = f (1) + 99 * 2F (1) ^ 2 = 296 ± 197 √ 2

Given that f (x) = loga ((1 + x) / (1-x)), (a > 0 and a is not equal to 1) find the definition field. Find the value range of X when f (x) > 0

one
Definition field (1 + x) / (1-x) > 0
That is, (x + 1) (x-1) < 0
x∈(-1.1)
two
f(x)>0
If a > 1, (1 + x) / (1-x) > 1
2x/(x-1)<0
x∈(0,1)
If 0 < a < 1, (1 + x) / (1-x) < 1
2x/(x-1)>0
X > 1 or X < 0
X ∈ (- 1,0) in combination with the domain

Given the number of culverts f (x) = loga (x + 1) - loga (1-x), a > 0 and a is not equal to 1, (1) find the definition field of F (x); (2) When a > 1, find X that makes f (x) > 0 Given the number of culverts f (x) = loga (x + 1) - loga (1-x), a > 0 and a is not equal to 1, (1) find the definition field of F (x); (2) When a > 1, find the value range of X that makes f (x) > 0

1、
If the true number is greater than 0, x + 1 > 0,1-x > 0
So define the domain (- 1,1)
2、
f(x)=loga[(x+1)/(1-x)]>0
a> 1, then loga (x) is an increasing function
And 0 = loga (1)
So (x + 1) / (1-x) > 1
(x+1)/(1-x)-1>0
2x(x-1)

The known function f (x) = the x power of 2 under the loga root minus 1, (a > 0 and a ≠ 1) ① find the definition domain of the function ② find the value range (process) of X with F (x) > 0

f(x)=loga(√(2^x-1))
√(2^x-1)>0
2^x-1>0
2^x>1=2^0
x>0
f(x)>0
√(2^x-1)>1
2^x-1>1
2^x>2^1
x>1

Find the definition field of function f (x) = LG (a ^ x-k * 2 ^ x) (a > 0 and a is not equal to 2)

A ^ x-k * 2 ^ x needs to be greater than 0
a^x>k*2^x
Because a > 0,2 > 0
So a ^ x > 0,2 ^ x > 0
therefore
(1) It holds when k * 2 ^ x is 0
k《0
(2) A ^ x > k * 2 ^ x > 0, that is, k > 0

It is known that f (x) is an even function, which is a subtractive function on [0, + ∞). If f (lgx) > F (1), the value range of real number x is () A. (1 10,1) B. (0,1 10)∪(1,+∞) C. (1 10,10) D. (0,1)∪(10,+∞)

∵ f (x) is an even function, which is a subtractive function on [0, + ∞),
‡ f (x) monotonically increases on (- ∞, 0),
From F (lgx) > F (1), f (1) = f (- 1)
Get: - 1 < lgx < 1,
∴1
10<x<10,
So the answer is C