The image of the function f (x) = X-2 parts ax + 3-A, and the image of the function y = g (x) and the image of the inverse function of y = f (x + 1) are symmetric with respect to the straight line y = X The analytic expression of inverse function of function y = f (x)

What does y = g (x) give you
If we find the analytic expression of the inverse function of y = g (x), the answer is y = (3 + x) / (x-a)

If the point (1,2) is not only on the image of the function f (x) = √ (AX + b), but also on the image of its inverse function f ^ - 1 (x), try to determine the analytic formula of F (x)

Replace (1,2) into the original function
Put (2,1) into the inverse function
By solving the equations, a = - 3, B = 7
Replace the original form

The function y = f (x) is the inverse function of function y = 2 ^ x, then what is the value of F (2)?

Let g (x) = 2 ^ X
Let g (x0 = 2 ^ x = 2
X=1
So g (1) = 2
Then f (2) = 1

Given that the function y = f (x + 1) - 2 is an odd function, y = f ˉ 1 (x) and y = f (x) are inverse functions of each other. If f (5) = 0, what is f ˉ 1 (4) equal to?

f(x+1)-2+f(1-x)-2=0
f(1+x)+f(1-x)=4;
Let x = 4, f (5) + F (- 3) = 4, because if f (5) = 0, then f (- 3) = 4;
Then f ˉ 1 (4) = - 3

If the point (1,2) is on both the image of the function y = √ (AX + b) and the image of its inverse function, then what are the values of a and B respectively

Point on the image of function y = √ (AX + b), 2 = √ (a + b), and on the image of inverse function, 1 = √ (2a + b), 4 = a + B.1 = 2A + B, a = - 3, B = 7

If the function y = f (x) is the inverse function of the function y = ax (a > 0 and a ≠ 1), its image passes through a point（ a. A), then f (x) =___ .

Let X be the inverse function of x = X（
a，a），
∴a=loga
a. That is, a = 1
2，
So the answer is: Log1
2x．

If the point (1,2) is both in the function y= On the graph of AX + B and its inverse function, then ab=______ .

Method 1: from the known: a + B = 2, that is, a + B = 4, and from y = ax + B to solve X: x = 1a (Y2 − b), then the inverse function of y = ax + B is y = 1a (x2 − b), ∵ point (1, 2) on the image of the inverse function

If the point (1.2) is not only on the image of the function y = √ ax + B, but also on its inverse function Thank you for the writing process

The point (1.2) is not only on the image of the function y = √ ax + B, but also on its inverse function, so (2,1) is also on the image of the original function

If the function y = f (x) is the inverse function of the function y = ax (a > 0, and a ≠ 1), and f (2) = 1, then f (x) = () A. log2x B. 1 2x C. log1 2x D. 2x-2

The inverse function of the function y = ax (a > 0 and a ≠ 1) is f (x) = logax,
And f (2) = 1, that is, loga2 = 1,
So, a = 2,
So f (x) = log2x,
Therefore, a

Find the inverse function of function y = {x + 1, x greater than or equal to 2 and 2 ^ X-1, X less than 2

Y = x + 1, x = 1-y. and x > = 2, so Y > = 3
y=2^x-1 x=log2(y+1).x

The image of the function f (x) = X-2 parts ax + 3-A, and the image of the function y = g (x) and the image of the inverse function of y = f (x + 1) are symmetric with respect to the straight line y = X The analytic expression of inverse function of function y = f (x)