If the function f (x) = the x power of a, the image crossing point (3,1) of the inverse function is equal to what

The image crossing point of the inverse function of the function f (x) = a ^ x (3,1)
Then the function f (x) = the image crossing point of a ^ x (1,3)
So 3 = a ^ 1
Then a = 3

Given the inverse function F-1 (x) = 2 of function f (x), how much is f (2) + F-1 (2) equal

f(-1)(2)=2²=4
Let F-1 (x) = 2 ^ x = 2
Then x = 1
f-1(1)=2
So f (2) = 1
So the original formula = 5

The inverse function of y = root (2x-x ^ 2) (1 ≤ x ≤ 2) is

2x-x^2
=-(x-1)^2+1
One

Y = inverse function of x square + X under root sign (x is greater than or equal to 1)

The original formula is as follows:
Y = radical (x ^ 2 + x)
So what
Y^2=X^2+X
(X+1/2)^2=Y^2+1/4
X = radical (y ^ 2 + 1 / 4) - 1 / 2
The inverse function is
Y = radical (x ^ 2 + 1 / 4) - 1 / 2
(where ^ 2 is the square)

Under y = radical (2x-x ^ 2), the inverse function is () with a known condition (1 is less than or equal to X and less than or equal to 2)

y=√(2x-x^2),1=

Y = inverse function of 2X-4 (x ≥ 2) under radical

y=√(2x-4)
The definition domain is x ≥ 2, and the range is y ≥ 0
y²=2x-4
∴ 2x=y²+4
∴ x=y²/2+2
That is, the inverse function is y = x 2 / 2 + 2, X ≥ 0

Find the inverse function of a function (1) y = 2x squared - 2 x ∈ * (- ∞ * 0) (2) y = the cube of root 2x + 1 x ∈ * (0 + ∞) known set a = {(x*

1)Y=2X^2-2
X=√1/2(Y+2)
The inverse function is y = √ 1 / 2 (x + 2)
2)y=(√2x+1)^3
x=(y^2/3-1)/2
The inverse function is y = (x ^ 2 / 3-1) / 2

What is the inverse function of the function f (x) = the square of X (x < equal to 0)

y=x^2
The inverse function is the exchange of X and y
That is, x = y ^ 2
Y = ± radical x
Because the original function x = 0

The inverse function of y = x squared (x ≥ 0) is?

Y = radical x (x > 0)

The square of 3x is 2x, and X is equal to

3x^2=2x
3x^2-2x=0
x(3x-2)=0
x=0,x=2/3

If the function f (x) = the x power of a, the image crossing point (3,1) of the inverse function is equal to what