Let f (x) = {arctan [1 / (x-1)], X not equal to 1,0, and x = 1. Let f (x) = {arctan [1 / (x-1)] be the limit at x = 1, and let f (x) be continuous at x = 1

Let f (x) = {arctan [1 / (x-1)], X not equal to 1,0, and x = 1. Let f (x) = {arctan [1 / (x-1)] be the limit at x = 1, and let f (x) be continuous at x = 1

The inverse function of the function y = 1 + Log1 / 2x is
y=2`(1-x) (x>o)
Y = 2 '(1-x) domain should be all real numbers R, which can be known from the original function range!
The function f (x) is a decreasing function on (0, + ∞). How to find the relationship between F (a2-a + 1) and f (34)?
∵ a2-a + 1 = (A-12) 2 + 34 ≥ 34 > 0 and the function f (x) is a decreasing function on (0, + ∞) ∵ f (a2-a + 1) ≤ f (34)
Let y = f (x) be a monotone odd function defined on (- ∞, + ∞). Is the inverse function monotone odd? Why?
This is the last question about inverse function
The original function and the inverse function are symmetric with respect to the straight line y = x in the rectangular coordinate system. Let y = f (x) and its inverse function be x = g (y) y = f (x) monotone =, x = g (y) monotone y = f (x) odd function, that is, f (0) = 0, f (x) = - f (- x) = > G (0) = 0, G (x) = - G (- x), so the original function y = f (x) is defined in (- ∞, + ∞)
Given that y = f (x) is an odd function and a decreasing function in the interval [0,4], then the relation between F (- π) and f (- 3) is
Y = f (x) is an odd function, so f (- π) = - f (π) f (- 3) = - f (3)
Y = f (x) is a decreasing function in the interval [0,4], so f (π) < f (3)
-F (π) > F (3), that is, f (- π) > F (- 3)
F (- π) is greater than f (- 3)
Because it is an odd function, f (- π) equals - f (π), f (- 3) equals - f (3)
And from zero to four, π is equal to more than three points, so f (π) is less than f (3)
So multiplying by - 1 is the opposite
It is known that A. B belongs to R, f (x) is an odd function, and f (2x) = (ax4 ^ x + A-2) / (4 ^ x + b). Find the inverse function of F (x) and its domain
First write the expression of F (x), f (x) = (AX2 ^ x + A-2) / (2 ^ x + b) f (x) is an odd function, then f (0) = 0, you can get a = 1, and f (x) = - f (- x), substituting f (x), f (- x), a = 1 respectively, you can get b = 1, so f (x) = (x2 ^ x-1) / (2 ^ x + 1) use y instead of F (x), you can get x = log2 [(1 + y) / (1-y)], from (1 +
If y = x ^ & # 178; + BX + 1 is an increasing function on X ∈ [1, + ∞), the value range of B is obtained
A:
The parabola y = x ^ 2 + BX + 1 is an increasing function when x > = 1
Because: parabola opening upward, axis of symmetry x = - B / 2
So: axis of symmetry x = - B / 2 = - 2
The increasing interval of y = (x-m) ² + n [M, + ∞) M = 1
y=(x-m)²+n=x²-2mx+m²+n
b=-2m
B
It is known that f (x) is a function defined on the set of real numbers, and its inverse function is f ^ - 1 (x)
If f ^ - 1 (x + a) and f (x + a) are reciprocal functions, and f (a) = a (a is a non-zero constant), then the value of F (2a) is?
In urgent need of answers, please solve the problem in detail, I will add a reward
Please give me a detailed solution
The inverse function of y = f ^ - 1 (x + a) is: x = f ^ - 1 (y + a)
y=f(x)-a=f(x+a)
f(2a)=f(a+a)=f(a)-a=a-a=0
In the mathematics of the first year of senior high school, f (x) = x & # 178; - (A-1) x + 5, X ∈ [0.5,1] is an increasing function, and the range of F (2) can be obtained
First, a ≤ 2 is calculated by using the previous conditions, and then f (2) is substituted to get f (2) = 11-2a
F (2) ≥ 7 can be calculated by a ≤ 2
If the minimum positive period of the function f (x) = sinax + cosax (a > 0) is 1, then a symmetry center of its image is ()
A. (−π8，0)B. （0，0）C. (−18，0)D. (18，0)
If f (x) = sinax + cosax = 2Sin (AX + π 4) t = 2 π a = 1, then a = 2 π, so f (x) = 2Sin (2 π x + π 4) Let f (x) = 0, then there is 2 π x + π 4 = 0x = - 18, that is, one of the symmetry centers is (- 18, 0), so C is chosen