When x > - 1, please find the range of F (x) = (x ^ 2-3x + 2) / x + 1

When x > - 1, please find the range of F (x) = (x ^ 2-3x + 2) / x + 1

F (x) = (x + 1) + 6 / (x + 1) - 5; simplify yourself, and then use the important inequality. I believe you can do this. The denominator is x ^ 2-3x + 2 = (x + 1) ^ 2-5 * (x + 1) + 6
Let f (x) be defined as R. for any real number x, y, f (x + y) = f (x) + F (y). When x > 0, f (x)
Let f (x) be defined as R. for any real number x, y, f (x + y) = f (x) + F (y). x> 0, f (x)
For example, y = 1 / X is monotonically decreasing on (- ∞, 0) and monotonically decreasing on (0, + ∞), but we can't think that y = 1 / X is monotonically decreasing on R. (1) proof: because f (x) has f (x + y) =... For any real number x, y
When x
f(x)=(x^2-3x+1)/(x+1)
=[(x^2+2x+1)-5(x+1)+5]/(x+1)
=[(x+1)^2-5(x+1)+5]/(x+1)
=(x+1)+5/(x+1)-5
Because x
If the domain of F (x) is r, f (θ + b) is satisfied for any real number a and B
The definition field of F (x) is R. for any real number a and B, f (θ + b) = f (θ) · f (b) is satisfied
Let f (x) > 1 and try to solve the inequality f (x + 5) > 1 / F (x) when x < 0
Give reasons
When x < 0, f (x) > 1 satisfies f (θ + b) = f (θ) · f (b) for any real number a and B
So f (- 1 + 0) = f (- 1) · f (0), - 11, so f (0) = 1
When x > 0, - x0, 0
The range of F (x) = Log1 / 2 (x ^ 2-3x + 2)
f(x)=log1/2(x^2-3x+2)
x^2-3x+2>0
The solution is x2
The domain of definition is (- ∞, 1) U (2, + ∞)
True number T = x ^ 2-3x + 2 = (x-3 / 2) ^ 2-1 / 4
When X2, t ∈ (0, + ∞)
∴log(1/2)t∈R
That is, the range of F (x) is r
It is known that the domain of definition of function f (x) is r, satisfying f (12) = 2, and for any real number m, n has f (M + n) = f (m) + F (n) - 1, when x ＞ - 12, f (x) > 0. (1) find the value of F (- 12); (2) prove that f (x) is a monotone increasing function on the domain of definition R
(1) For any real number m, N, f (M + n) = f (m) + F (n) - 1, let m = n = 0, then f (0) = f (0) + F (0) - 1, that is, f (0) = 1, let m = 12, n = - 12, then f (0) = f (12) + F (- 12) - 1 = 1, since f (12) = 2, then f (- 12) = 0; (2) it is proved that when x ＞ - 12, f (0) = f (12) + F (- 12) - 1 = 1, then f (- 12) = 0
Find the range; F {x} = x & # 178; - 1 of X & # 178; - 3x + 2
x≠±1
f(x)=(x-1)*(x-2)/[(x+1)*(x-1)]
=(x-2)/(x+1)
=1-[3/(x+1)]
f(x)≠1
It is known that the domain of definition of function f (x) is R. for any real number m, N, f (1 / 2) = 2, and f (M + n) = f (m) + F (n) - 1, when x > - 1 / 2, f (x) > 0
Find the value of (1) f (- 1 / 2)
(2) To prove that f (x) increases monotonically over the domain R
The key point is the second question, the problem of finding the domain of abstract function
(1) From the meaning of the question, we can know: F (1 / 2) = f (1 / 4 + 1 / 4) = f (1 / 4) + F (1 / 4) - 1 = 2
F (1 / 4) = 1.5
f(1/4)=f(1/2-1/4)=f(1/2)+f(-1/4)-1
f(-1/4)=f(1/4)-f(1/2)+1=0.5
f(-1/2)=f(-1/4-1/4)=f(-1/4)+f(-1/4)-1=0
(2) In the domain R, let a > b, let a = B + C, then there must be C > 0
f(a)=f(b+c)=f(b)+f(c)-1
Then f (a) - f (b) = f (c) - 1
And f (c) = f (C-1 / 2 + 1 / 2) = f (C-1 / 2) + F (1 / 2) - 1 = f (C-1 / 2) + 1
Because C > 0. So C-1 / 2 > - 1 / 2. That is f (C-1 / 2) > 0
therefore
f(a)-f(b)=f(c-1/2)+1-1=f(c-1/2)>0
In the domain of X, when a > b, f (a) > F (b)
So f (x) is monotonically increasing
(2) Arbitrary XX = = > Y-X > 0 = = > y-x-1 / 2 > - 1 / 2 = = > F (y-x-1 / 2) > 0
So f (y) > F (x)
Given f (x + 2) = x & # 178; + 3x-1, the expression of y = f (X-2) and its range
Let T-2 = x + 2, and then let T-2 = t-4f (T-2) = f (x + 2) = t-4f (t-4f (T-2) and let t-4f (T-2) = f (x (x + 2) \\35\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\it's not easy
f(x+2)=x²+3x-1 ＝(x+2)²-(x+2)-3
So, f (x) = x & # 178; - x-3
Therefore, f (X-2) = (X-2) &# 178; - (X-2) - 3 = x & # 178; - 3x - + 3
Because f (X-2) = (x-3 / 2) &# + 3 / 4
So, the range of Y is Y > = 3 / 4
If f (lgx) > F (1), then the value range of real number x is ()
A. （110，1）B. （0，110）∪（1，+∞）C. （110，10）D. （0，1）∪（10，+∞）
∵ f (x) is an even function, it is a decreasing function on [0, + ∞), and∵ f (x) increases monotonically on (- ∞, 0). From F (lgx) > F (1), f (1) = f (- 1), we get: - 1 < lgx < 1, ∵ 110 < x < 10, so the answer is C