Inverse function of y = 1-radical (1-x ^ 2) (- 1 less than or equal to x less than 0)

Y = 1-radical (1-x ^ 2)
1-x^2=(1-y)^2
x^2=1-(1-y)^2
x^2=2y-y^2
Because: - 1 ≤ x

Find the inverse function of y = LG (x + root x square - 1)! Detailed steps! Thank you!

Y = LG [x + radical (x ^ 2-1)]; the following inverse solution is X:
10 ^ y = x + radical (x ^ 2-1);
10 ^ Y-X = root (x ^ 2-1); square both sides
100^y-2*10^y*x+x^2=x^2-1;
2*10^y*x=100^y+1;
x=(100^y+1)/(2*10^y);
The standard form is
f^(-1)(x)=(100^x+1)/(2*10^x);

What is the inverse function of 1 + x square under y = 1 + radical

Exchange X and y of the original function
therefore
Inverse function of 1 + y square under x = 1 + radical
Come out again
Pay attention to the definition domain, the value domain

How to find the inverse function of x square + X under y = radical?

First of all, let 1 determine the range of Y, so y is greater than or equal to 0. If x squared + X is nonnegative, then x is greater than or equal to 0 or X is less than or equal to negative 1
Let's say that the square of y = x square + X,
At this time, the formula should be considered, which is one of the commonly used methods;
If 1 / 4 is added to both sides of equation 2, the square of Y + 1 / 4 = (x + 1 / 4) is obtained from the formula on the right
If we open the root sign again, we can get x = under the root sign (the square of Y + 1 / 4) - 1 / 2
Then change YX, indicate the range, OK

Let x + 3 + be equal to x + 3

y = 4+sqrt(3+x)
(y-4)^2=3+x
y^2-8y+13=x
The inverse function g (y) = y ^ 2-8y + 13 (Y > = 4 + sqrt (6))

Y = x square - 2x + 3, (x

Inverse function: x = √ [(Y-1) 2] + 2] inverse function: x = √ [(Y-1) 2] = ± √ 2 [the original function range is y ≥ √ 2] inverse function: x = √ [(Y-1) 2] + 2] = (Y-1) mm2 + 2 (Y-1)] = x? - 2y-1 = ± √ (x? - 2) [because the original function definition domain x ≤ 1, the value range of inverse function is

Find the function y = root (x square - 2x + 3) (x

Let y = the square of 1-x under the root sign; (- 1)

-1

Y = the square of X + 1, X is greater than or equal to - 1?

Because x > = - 1
So x + 1 > = 0
So x + 1 = √ y
x=√y-1
So the inverse function is y = √ X-1 (x > = 0)

X squared minus 3x-2 equals 2x? I don't know whether to subtract 3x minus 2 or subtract (3x-2)

Suppose x squared minus 3x-2 equals 2x
x^2-3x-2=2x
x^2-5x-2=0
x^2-5x-25/4+17/4=0
(x-5/2)^2+17/4=0
Because (X-5 / 2) ^ 2 > = 0
So the original (X-5 / 2) ^ 2 + 17 / 4 > 0
Not established
So x squared minus 3x-2 is not equal to 2x

Inverse function of y = 1-radical (1-x ^ 2) (- 1 less than or equal to x less than 0)