Inverse function of y = x ^ 2-2 * x + 5

A:
y=x^2-2x+5=(x-1)^2+4>0
So:
(x-1)^2=y-4>=0
X-1 = √ (y-4) or X-1 = - √ (y-4)
So:
X = 1 + √ (y-4) or x = 1 - √ (y-4)
So:
The inverse function corresponds to two branches y = 1 + √ (x-4) or y = 1 - √ (x-4)

The inverse function of y = arcsin (5-x) / 2 can't we just remove arc?

If y = arcsinx, arc can be removed directly. Or the inverse function of F ((5-x) / 2) = arcsin (5-x) / 2) can remove arc directly
In fact, no matter what the problem is, the standard solution should be to find the expression of X expressed by Y, that is, x = f (y)
The correct way to solve this problem is to put sin function on both sides at the same time,
Siny = sin (arcsin (5-x) / 2)
Then, according to the definition of inverse function, sin (arcsinx) = X
So siny = (5-x) / 2
So x = 5-2siny
So the inverse function is x = f (y) = 5-2siny, and Y is used to represent the dependent variable, so it becomes y = 5-2sinx

Finding the inverse function of y = x ^ 5 + 2

y-2=x^5
x=(5)√(y-2)
y∈R
Inverse function of y = x ^ 5 + 2
y=(5)√(x-2)
x∈R

Finding the inverse function of y = 2 ^ (x + 5)

y=2^(x+5)
lny=(x+5)ln2
lny/ln2=x+5
x=lny/ln2-5
So the inverse function is:
f(x)=lnx/ln2-5.

A high one inverse function problem arcsin(-sinx)+arccos(-sinx)=π/2 Find the value range of X,

The domain of arcsin T & arccos t is [- 1,1]
-1

An inverse function problem in senior one`~ Y = ax + 1 and y = (x / 3) + B are inverse functions of each other There must be a process ~ ~ otherwise you can't understand ~ ~ 3Q~~

y=ax+1.(1)
y=(x/3)+b.(2)
By changing y and X in (1), x = ay + 1... (3)
y=x/a-1/a ...(4)
(2) (4) equal, so a = 3, B = - 1 / 3

A question about high one inverse function, master come Given the function f (x) = (x-a + 1) / (x-a), and the symmetry center of the image of its inverse function is m (m, 3), what is the value of a? Please talk about the specific process, thank you! Wolf, ask you why the center of symmetry of Y-1 = 1 / (x-a) is (a, 3)

A = 3, this is a V-shaped function, which is a deformation of y = x + 1 / x, which can be transformed into Y-1 = 1 / (x-a). Therefore, the symmetry center of the function is (a, 3) and the original function is symmetric about the origin, and the symmetric center of the inverse function is (3, a), so a = 3

High one-way inverse function Inverse function of F (x) = log2 [(x ^ 2 + 1) ^ 1 / 2 - x] Note: in brackets is a root sign. Inside is x square + 1. Then a root sign minus X

The number I worked out is
f(2)=log(2)*((((2^2)+1)^(1/2)) -2)
f =0.1636

The problem of inverse function in high school The symmetric center of the inverse function y = (A-X) / (x-a-1) is (m, 3). Find the value of a, M I know the answer, but I hope the prawns will give the detailed process! Thank you very much! Hope that the prawns who answered the question will come back from time to time to see if I have any questions to add! My friend on the first floor understood it, but I didn't know how the second line (2m, 6-A) and (2m + A / A + 1,6) came from.

y=(a-x)/(x-a-1)
Can be converted into
y+1=-1/(x-a-1)
It can be seen that y = - 1 / X translates according to the vector (a + 1, - 1)
The symmetry center of y = - 1 / X is (0,0)
The symmetry center of Y + 1 = - 1 / (x-a-1) is (a + 1, - 1)
It is also known that the symmetry center of its inverse function is (m, 3)
The symmetry center of Y + 1 = - 1 / (x-a-1) is (3, m)
That is, a + 1 = 3 M = - 1
∴a=2 m=-1

What is the inverse of the function y = 2X-4?

Y=2x-4
==>x=(y+4)/2
==>Inverse function y = x / 2 + 2

Inverse function of y = x ^ 2-2 * x + 5