It is proved that the inverse function of the function y = (1-x) / (1 + x) (x is not equal to - 1) is the function itself

It is proved that the inverse function of the function y = (1-x) / (1 + x) (x is not equal to - 1) is the function itself

Multiply both sides by 1 + X to get (1 + x) y = 1-x,
Y+XY=1-X
XY+X=1-Y
X(1+Y)=1-Y
X=(1-Y)/(1+Y)
The range of y = - 1 + 2 / (1 + x) is (- ∞, - 1) ∪ (- 1, + ∞)
That is, y is not equal to - 1
So x = (1-y) / (1 + y) (y is not equal to - 1)
Get the certificate

Find the inverse function of the function y = 1 / x + 5 (x is not equal to - 5) The inverse function of the function y = 1 / (x + 5) (x is not equal to - 5) is found without a bracket

Exchange X and y
X = 1 / (y + 5)
Where x = (x) is not equal to 0
The inverse function of the function y = 1 / x + 5 (x is not equal to - 5) is y = (1 / x) - 5, X is not equal to 0

For the image of the inverse function of the function f (x) = loga (x-1) (a > 1, a ≠ 10), then a = ()

According to the definition of inverse function, if f (x) = loga (x-1) passes through point (4,1), then f (4) = 1, so loga (4-1) = 1
A=3

If we know the image crossing point (4,0) of the function f (x) = loga (x-k), and its inverse function y = the image crossing point (1,7) of f ^ - 1 (x), then f (x) is A. Increasing function B. decreasing function C. increasing then decreasing D decreasing first then increasing

B. Subtraction function

Given that the inverse function of the function y = f (x) is y = 1 + loga (1-x) (a > 0, and a ≠ 1), then the image of function y = f (x) must pass through the point () Why?

(1,0) because the inverse function must pass through the point (0,1), and then exchange the position of X and y to get the point that the original function must pass through

Let a > 0, and a ≠ 1, f (x) = loga (x + √ (x ^ 2-1)) (x ≥ 1), find the inverse function f ^ - 1 (x) of function f (x)

a^y=x+√(x^2-1)
a^y-x=√(x^2-1)
square
a^(2y)-2x*a^y+x^2=x^2-1
a^(2y)-2x*a^y=-1
x=[a^(2y)+1]/(2a^y)
The inverse function is y = [a ^ (2x) + 1] / (2a ^ x)

The known function f (x) = loga (A-A ^ x) (0 No one can do it?

One
y=loga(a-a^x),
a-a^x=a^y,
a^x=a-a^y
x=loga(a-a^y),
x. Y interchange,
f^-1(x)=loga(a-a^x)；
Definition domain: A-A ^ x > 0 = = > a > A ^ x = = > x > 1,
Two
F ^ - 1 (x ^ 2-2) 2, or x0
x>√2,x2.

When x ∈ [- 2,1], the value range of the function f (x) = x2 + 2x-2 is () A. [1，2] B. [-2，1] C. [-3，1] D. [-3，+∞）

The function f (x) = x2 + 2x-2 = (x + 1) 2-3, and the symmetry axis of parabola is x = - 1
Because x ∈ [- 2,1], when x = - 1, the minimum value of the function is f (- 1) = - 3
Because 1 is far away from the symmetry axis, when x = 1, the function obtains the maximum value f (1) = 1 + 2-2 = 1
So the value range of the function is [- 3,1]
Therefore, C

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

yx^2+yx+y=2x^2+2x+5
(y-2)x^2+(y-2)x+y-5=0
The equation must have △≥ 0
(y-2)^2-4(y-2)(y-5)≥0
(y-2)(y-6)≥0
Y ≥ 6 or Y ≤ 2

If f (x) = (1 / 2) x is known and its inverse function is g (x), then the domain of G (2x-6) is

The definition domain of inverse function is the range of original function, the range of F = 0.5x is infinite, so the range of 2x-6 is infinite, and the range of X is also infinite... So the definition domain of G (2x-6) is (- infinite, infinite)