If the inverse function of y = 3 ^ x is y = f (x), then f (1 / 9) =?

If the inverse function of y = 3 ^ x is y = f (x), then f (1 / 9) =?

y=3^x
x=log3 y
When x and y are exchanged, we get
y=log3 x
f(x)=log3 x
∴f(1/9)=log3 (1/9)=log3 (3^(-2))=-2*log3 (3)=-2

If the function y = 1 + 2 ^ x, then its inverse function is

2^x=y-1,(2^x>0,1+2^x>1)
x=log2(y-1)
y=log2(x-1),(x>1)

The inverse function of the known function y = 2 / 3 √ 9-x ^ 2 (- 3 ≤ x < 0) is_______ Write the process

√(9-x^2)=3y/2
square
9-x^2=9y^2/4
x^2=9-9y^2/4
If x < 0, x = - √ (9-9y ^ 2 / 4)
So y = - √ (9-9x ^ 2 / 4), 0 ≤ x < 2

Let y = f (x + a), y = F-1 (x + a) be inverse functions of each other to find f (x)

Because y = F-1 (x + a), f (y) = x + A, x = f (y) - A
So the inverse function y = F-1 (x + a) is y = f (x) - A
So f (x) = y + a

It is known that the inverse function of the function f (x) = 3 ^ X-1 is y = f ^ - 1 (x), G (x) = log with 9 as the base (3x + 1) (1) If f ^ - 1 (x) ≤ g (x), find the value range D of X (2) Let H (x) = g (x) - (1 / 2) f ^ - 1 (x). When x belongs to D, find the range of H (x)

(1) Y = 3 ^ x-1y + 1 = 3 ^ x, i.e. x = log (3) (y + 1), so the inverse function is y = log (3) (x + 1) = log (9) (x + 1) ^ 2 (x > - 1) so (x + 1) ^ 2 ≤ 3x + 1, the solution d = [0,1] (2) H (x) = log (9) (3x + 1) - log (9) (x + 1) = log (9) [3-2 / (x + 1)] when 0 ≤ x ≤ 1, - 2 / (x + 1) ∈ [- 2, - 1] 3-2 / (x + 1)

How to find the inverse function of y = f (x + 1) Like this, there's an F Then how do you calculate the inverse function? How do you know x =? F is shrimp Who can make it clear

What is the number of symbols X and f that means a function
Let y = x, then f (x + 1) = X
f(x+1)=(x+1)-1=x
So f (x) = X-1
y-1=x-1
So the inverse function of y = f (x + 1) is f (x) = Y-1

If f (x) has an inverse function, y = f (x + 1) is over charged (3,1), then y = f ^ - 1 (x) must cross the point?

f(3+1)=f(4)=1
Then the inverse function passes (1,4)

If the function y = f (x) has an inverse function and f (3) = 0, then the image of function F-1 (x + 1) must pass through the point () A. （2，0） B. （0，2） C. （3，-1） D. （-1，3）

Analysis: ∵ the function y = f (x) has an inverse function and f (3) = 0,
Then the graph of function f (x) passes through point (3,0),
The graph of the inverse function F-1 (x) of the function f (x) passes through point a (0, 3),
Then F-1 (0) = 3,
Then the graph of function F-1 (x + 1) must cross the point (- 1,3)
Therefore, D

Find the inverse of the following function: y = 2 ^ x ^ 2-2x + 3 (x > = 1)

x²-2x+3=log2(y)
(x-1)²=log2(y)-2
x>=1
Then X-1 = √ [log2 (y) - 2]
x=1+√[log2(y)-2]
x>=1
Then x? - 2x + 3 > = 2
y>=2²=4
So the inverse function is y = 1 + √ [log2 (x) - 2], x > = 4

Find the inverse function y = x ^ 2-2x + 3 (x > 1)

Inverse function y = x ^ 2-2x + 3 (x > 1)
y=(x-1)²+2;＞2;
∴（x-1）²=y-2;
x-1=√(y-2);
x=√(y-2)+1;
So the inverse function is y = √ (X-2) + 1 (x > 2)
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