Given the function f (x) = loga (AX radical x) (a > 0, a is not equal to 1 is a constant) (1) find the definition domain of function f (x) (2) if a = 2, try to define it according to monotonicity Given the function f (x) = loga (AX radical x) (a > 0, a is not equal to 1 is a constant) (1) find the definition domain of function f (x) (2) If a = 2, try to determine the monotonicity of function f (x) according to the definition of monotonicity (3) If the function y = f (x) is an increasing function, find the value range of A

1. Function y = - 1-x (x) under radical sign

1.y=-√(1-x) √(1-x)=-y 1-x=y^2 x=1-y^2 y^(-1)=1-x^2,x

If the function y = f (x) is the inverse function of the function y = a ^ x (0 < A is not equal to 1) and its image passes through the point (root sign a, a), then the function y = - f (mx-4) is an increasing function on the interval (2, infinity), then the value range of positive number m is as follows:

Y = f (x) is the function y = a ^ x (0 then a ^ a = a ^ V),
a=½
Y = - f (mx-4) = - (half) ^ (mx-4) is an increasing function on the interval (2, infinity)
Then: (1 / 2) ^ (mx-4) is a decreasing function on the interval (2, infinity)
Mx-4 is an increasing function on the interval (2, infinity)
So m > 0

Let f (x) = loga (x + b) (a > 0, a ≠ 1) and its inverse function (2, 8), then a + B is equal to () A. 6 B. 5 C. 4 D. 3

The image crossing point (2,1) of the function f (x) = loga (x + b) (a > 0, a ≠ 1) and the image crossing point (2,8) of its inverse function,
be
loga(2+b)＝1
loga(8+b)＝2 ，
Qi
2+b＝a
8 + B = A2, a = 3 or a = - 2 (round), B = 1,
∴a+b=4，
Therefore, C

If f (x) = 3 ^ (2x-1), the inverse function is

y=3^(2x-1),y>0
log3(y)=2x-1
x=(1/2)log3(3y)
Therefore, the inverse function is f ^ - 1 (x) = (1 / 2) log3 (3x), x > 0

Find the inverse function of F (x + 1) = x2 + 2x + 3

The inverse function of F (x) = x 2 + 2x (x ≤ - 1) I know the answer, but I don't know how to get it,

y=x²+2x
=x²+2x+1-1
=(x+1)²-1
(x+1)²=y+1
X + 1 = √ (y + 1) or x + 1 = - √ (y + 1)
X = √ (y + 1) - 1 or x = - √ (y + 1) - 1
Inverse function
Y = √ (x + 1) - 1 or y = - √ (x + 1) - 1

The function f (x) = x? - 2x (x

y=x^2-2x=(x-1)^2-1
y+1=(1-x)^2
Because of X

2x-1 (x

Your f (x) is a piecewise function, right
The definition of inverse function is called inverse function
Primitive function and inverse function are symmetric about y = x (f (x) = y. inverse function (y) = x)
Then the inverse function (- 3 / 4) is f (x) = - 3 / 4
2x-1 = - 3 / 4 x = 1 / 8 (condition x = 0 takes x = 1 / 2)
Therefore, it is d

If the function f (x) = x 2 + 2x (x ≥ 0), then the domain of inverse function is defined

The definition domain of inverse function is the value range of function
f(x) = x^2+2x = x^2+2x+1 -1 = (x+1)^2 -1 => f(x) >= 0
Let f (y) be the inverse function of F (x), that is, f (x) = Y > = 0
So the definition domain of inverse function [0, infinite)