The range of the function y = 1 / 2 of X-1 (x > 1) is?

The range of the function y = 1 / 2 of X-1 (x > 1) is?

x>1
x-1>0
1/(x-1)>0
y>0

14: Function f (x) = | (2 ^ x) - 1 |, if AF (b), then the following four formulas hold
A.a

When - ∞

If f (x) = cosx / 2, then
Why f (- x) = f (x)

Because cosx is an even function, X becomes negative and the value of the function remains unchanged

It is known that f (x) is a function defined on R, and f (1) = 1. For any x ∈ R, the following two formulas hold: (1) f (x + 5) ≥ f (x) + 5; (2) f (x + 5) ≥ f (x) + 5
It is known that f (x) is a function defined on R, and f (1) = 1. For any x ∈ R, the following two formulas hold: (1) f (x + 5) ≥ f (x) + 5; (2) f (x + 1) ≤ f (x) + 1. If G (x) = f (x) + 1-x, find the value of G (6)

According to equation 2), we can get
f(x+5)≤f(x+4)+1 ≤f(x+3)+2 ≤f(x+2)+3≤f(x+1)+4≤f(x)+5
According to equation 1, we get
f(x+5) = f(x)+5
f(6)=f(1)+5 = 6
g(6)=f(6)+1-6 = 1

If f (x) is a differentiable function defined on R, and f '(x) > F (x), for any positive real number a, then the following formula holds true
A、f(a)＜eaf(0) B 、f(a)＞eaf(0) C、f(a)＜f(0)/ea D、f(a)＞f(0)ea

A:
f'(x)>f(x)
If f '(x) - f (x) > 0, multiply both sides by e ^ (- x) > 0
f'(x)*e^(-x)-f(x)*e^(-x)>0
So: [f (x) e ^ (- x)] '> 0
So: [f (x) / e ^ x] '> 0
So: F (x) / e ^ x is an increasing function
So for any positive real number a > 0
So: F (a) / e ^ a > F (0) / e ^ 0 = f (0)
So: F (a) > F (0) e ^ A
Option B and option D seem to be the same? Choose f (0) times e to the power of A

Let the domain of definition of function f (x) be d. if for any x 1 ∈ D, there is a unique x 2 ∈ d corresponding to it and (f (x 1) + F (x 2)) / 2 = C (C is a constant), then the mean value of function y = f (x) on D is called C. The following four functions are given
①y=x
②y=|x|
③y=x^2
④y=1/x
⑤y=x+ 1/x
Write out the sequence numbers of all functions whose mean value is 2
Only one is right?

1. Yes
2. If x is greater than 2, there is no solution
3. There is no solution when the same as above is greater than root 2
5. There is no correspondence when x = 2
4. There is no solution when x = 1 / 2

The function f (x) is differentiable in the open interval (a, b), and f '(x) is monotone in (a, b). This paper proves that f' (x) is continuous in (a, b)

Let f '(x) monotonically increase. Let C ∈ (a, b). F' (c) = LIM (H → 0 -) {[f (c + H) - f (c)] / h}. From Lagrange's theorem: there exists ξ ∈ (c + H, C). F (c + H) - f (c) = f '(ξ) H (when h < 0, ξ < C)

If the derivative of F (x) is f '(x) = - x (x + 1), then G (x) = f (logax)) (0

Because f '(x) = - x ^ 2-x, according to the derivation principle of compound function: G' (x) = [- logax (logax + 1)] * 1 / (x * ln a) g '(x) = [- logax (logax + 1)] * 1 / (x * ln a) ≤ 0 ∵ 0 < a < 1 ∵ LNA < 0 ∵ x > 0, logax (logax + 1) ≥ 0, we get: ① logax ≥ 0 = = > 0 < x ≤ 1 or: ② logax ≤ - 1 = = > x ≥

Let f (x) satisfy (x Λ 2-1) f '(x) > 0, then the monotone increasing interval of F (x) is_____________ .
Online urgent! Help! Seek process and answer! Thank you

(- ∞, - 1), (1, ∞) must be right!

F (x) = 1 / 3x ^ 3-ax ^ 2-3a ^ 2X-4 on (3, + infinity) is the range of increasing function to find a
f(X)=1/3X^3-aX^2-3a^2X-4
f'(X)=(X-3a)(X+a)
Let f '(x) = 0
→: x = 3A or: x = - A
When x = 3A > - A: a > 1
When x = - a > 3a, a

f(X)=1/3X^3-aX^2-3a^2X-4
f'(X)=(X-3a)(X+a)
Let f '(x) = 0
→: x = 3A or: x = - A
When x = 3A > - A: a > 1
When x = - a > 3a, a