F (x) = / 2x-1 / - / 2x + 1 /, judge the parity function

F (x) = / 2x-1 / - / 2x + 1 /, judge the parity function

f(x)=|2x-1|-|2x+1|
be
f（-x）=|-2x-1|-|-2x+1|
=|-（2x+1）|-|1-2x|
=|2x+1|-|1-2x|
=-f（x）
So the function f (x) = | 2x-1 | - | 2x + 1 | is odd

Given the function f (x) = 2x + X / A and f (1) = 1, it is necessary to find the value of real number a and judge the parity of function f (x)
2) Is the function f (x) an increasing or decreasing function on (1, + ∞) and proved by definition

F (x) = 2x + A / X for AF (1) = 12 + A / 1 = 1A = - 1F (x) = 2x-1 / X for parity it is easy to know that the domain of definition is symmetric about the origin f (- x) = 2 (- x) - 1 / (- x) = - 2x + 1 / x = - (2x-1 / x) = - f (x), so it is odd function for monotonicity, let X1 > x2 > 1F (x1) - f (x2) = (2x1-1 / x1) - (2x2-1 / x2) = 2 (x1-x2) - (1 / x1-1 / x2)

Given the function f (x) = (2x-1) / (x + 1) (1) find the domain of definition of F (x); (2) prove that the function f (x) = (2x-1) / (x + 1) in
Given function f (x) = (2x-1) / (x + 1)
(1) Find the domain of F (x);
(2) It is proved that the function f (x) = (2x-1) / (x + 1) is a monotone increasing function on [1, + ∞)
Please, thank you

Given function f (x) = (2x-1) / (x + 1)
(1) Find the domain of F (x);
X+1≠0
X≠-1
(2) It is proved that the function f (x) = (2x-1) / (x + 1) is a monotone increasing function on [1, + ∞)
f(x)=(2x-1)/(x+1)=2-3/(x+1)
Let 1 ≤ x1

It is known that the minimum value of even function f (x) defined on R is 1, and when x > = 0, f (x) = e ^ x + A, where e is the base of natural logarithm
(1) Find the analytic expression of function f (x)
(2) If the function f (x) = f (x) - BX ^ 2 has exactly two different zeros, find the value of B

(1) We can judge that x > = 0, f (x) is a monotonically increasing function, because f (x) is an even function,
x=0,f(x)=e^x x

For the even function f (X-2) defined on R, when x ＞ - 2, f (x) = ex + 1-2 (E is the base of natural logarithm), if there exists K ∈ Z such that the real root of the equation f (x) = 0 x0 ∈ (k-1, K), then the value set of K is ()
A. {0}B. {-3}C. {-4，0}D. {-3，0}

The graph of ∵ even function f (X-2) is symmetric about Y-axis ∵ the image of function y = f (x) is symmetric about x = - 2 ∵ when x ∵ 2, f (x) = ex + 1-2 ∵ f (x) = ex + 1-2 increases monotonically at (- 2, + ∞), and f (- 1) < 0, f (0) = E-2 > 0. From the existence theorem of zeros, we can see that the function f (x) = ex + 1-2 has zeros at (- 1, 0). From the symmetry of function image, when x