The inverse function of F (x) = (1 + x ^ 2) sgnx is?

The inverse function of F (x) = (1 + x ^ 2) sgnx is?

Let f (x) be written as a piecewise function, f (x) = 1 + x ^ 2, x > 0, f (x) = 0, x = 0, f (x) = - (1 + x ^ 2), x < 0. When x > 0, let y = f (x) = 1 + x ^ 2, then x = √ (Y-1), Y > 1. When x < 0, let y = f (x) = - (1 + x ^ 2), then x = - √ (- Y-1), y < 1. Note that f (- 1) (x) is the inverse function of F (x), then f (- 1) (x) = √ (x-1), x > 1F (

F (x) = x + k of a over (1,3), its inverse function over (2,0)
I calculate roughly a = 2, k = 1, the calculation is wrong

f(1)=3
So a + k = 3
If the inverse function is over (2,0), then f (x) is over (0,2)
SO 2 = a ^ 0 + k = 1 + K
So you did the right thing

The inverse function of F (x) = a ＾ x-3 + 1 (a > 0 and a ≠ 1)
Which fixed point does this inverse function pass through?

(1,0)

If the function f (x) = 3 ^ x, its inverse function is f ^ - 1 (x), and f ^ - 1 (27) = a + 2, find the value of A

f^-1(27)=a+2
That is to say, when x = a + 2, y (x) = 27, it is brought into the equation
3 ^ (a + 2) = 27, then a = 1