It is known that there is an inverse function on R defined by F (x), and f (9) = 8. If y = f (x + 1) inverse function is y = f ^ - 1 (x + 1), then f (2008)= And f (9) = 18, sorry for the wrong number!

It is known that there is an inverse function on R defined by F (x), and f (9) = 8. If y = f (x + 1) inverse function is y = f ^ - 1 (x + 1), then f (2008)= And f (9) = 18, sorry for the wrong number!

The inverse function of y = f (x + 1) is y = f ^ - 1 (x) - 1
f^-1(x)-1=f^-1(x+1)
f^-1(x)=f^-1(x+n)+n
Because f (9) = 18, that is, f ^ - 1 (18) = 9
If f ^ - 1 (x) = 2008, there is f ^ - 1 (x + n) + n = 2008, and N = 1999
F ^ - 1 (x + 1999) = 9
F ^ - 1 (x) is a monotone decreasing function with x + 1999 = 18, that is, x = - 1981
F ^ - 1 (- 1981) = 2008, that is, f (2008) = - 1981

Inverse function f (8) = 9 y = f (x + 1) y = f ˉ (x + 1) find f (2008)=

From y = f ˉ (x + 1), y = f (x) - 1 = f (x + 1) can be obtained
f（2008）=f（2007）-1=f（2006）-2=…… =f（8）-2000=9-2000=1991...
I haven't done these questions for a long time

Finding the inverse function (process) of the function y = π / 2 + arcsinx, X ∈ [- 1,1]

-π/2

The inverse function of the function y = 3 ^ (x ^ 2-1) x ∈ [- 1,0) is

Therefore, the inverse function of y = 3 ^ (x ^ 2-1) x ∈ [- 1,0) is y = √ log3 x ∈ (1 / 3, + ∞) the domain of definition of the original function is the range of the inverse function

If f (3) = 0, then the image of the inverse function of F (x + 1) must cross the point. A (2,0) B (0,2) C (3, - 1) d (- 1,3)

If f (3) = 0, then (3,0) must be on the original function. Therefore, when x = 2 on a new function, the point (0,2) must be on its inverse function
So choose B

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

The function f (x) = 2 ^ x is an exponential function, and its inverse function is a logarithmic function. In the function f (x) = 2 ^ x, the solution of X is x = log Ψ f (x)
That is, y = f ^ - 1 (x) is: y = log Ψ x, then f ^ - 1 (3) = log Ψ 3
Fill in log Ψ 3

If the function y = f (x) is the inverse function of the function y = 3x, then f (1) 2) The value of is () A. -log23 B. -log32 C. 1 Nine D. Three

∵ the function y = f (x) is the inverse function of the function y = 3x,
∴y=f（x）=log3x，
∴f（1
2）=log31
2=-log32
Therefore: B

It is known that the function y = f (x) has an inverse function y = f ^ negative 1 (x). If the image of the function y = f (x + 1) passes through point (3,1), then the image of function y = f ^ minus 1 (x) must pass through It is known that the function y = f (x) has an inverse function y = f ^ negative 1 (x). If the image of function y = f (x + 1) passes through point (3,1), then the image of function y = f ^ negative 1 (x) must pass through the point? Emergency

The image of function y = f (x + 1) passes through point (3,1), that is, f (4) = 1
So f ^ minus 1 (1) = 4, so the image of function y = f ^ minus 1 (x) must pass through (1,4)

If the inverse function of function f (x) is f ^ (- 1) (x), and y = f ^ (- 1) (- x + 2) is over (- 1,2), then y = f (x-1) is over a fixed point If the inverse function of function f (x) is F-1 (x), and y = F-1 (- x + 2) is over (- 1,2), then y = f (x-1) is over a fixed point Let y = ax, f ^ (- 1) (x) be x = ay, f ^ (- 1) (- x + 2) be - x + 2 = ay, and from over (- 1,2) to 3 = 2A, then f (x-1) i.e. y = (x-1) a always crosses (3,3) point, The answer is (- 3, - 3)

A:
The inverse function F-1 (x) of y = f (x), y = F-1 (- x + 2), passes through the point (- 1,2), i.e., y = F-1 (1 + 2) = F-1 (3) = 2
So: the original function y = f (x) passes through point (2,3)
Let X-1 = 2, then y = 3
So: x = 3, y = 3
So: y = f (x-1) passes through the fixed point (3,3)
I think the answer is wrong
The answer process of (- 3, - 3) should be as follows: y = F-1 (- x + 2) goes through point (- 1,2)
Let - x + 2 = - 1
The solution is: x = - 3
Inverse function y = F-1 (x) passing through point (- 3,2)
Original function passing through point (2, - 3)
In y = f (x-1), let X-1 = 2, x = 3, y = - 3
The result is (3, - 3), and it can't be (- 3, - 3)
Moreover, y = F-1 (- x + 2) is to find the inverse function first, and then put - x + 2 into it

The inverse function of y = f (x + 1) is y = F-1 (x + 1) f (0) = 1 F-1 (2) =? Such as the problem solving process The answer is - 1 Yeah

y=f-1(x+1), x+1=f(y),x=f(y)-1
By exchanging x, y, we get:
The inverse function of y = F-1 (x + 1) is y = f (x) - 1
That is, f (x + 1) = f (x) - 1
f(x)=f(x+1)+1
f(-1)=f(0)+1=2
So F-1 (2) = - 1