Is the inverse function of the function y = (x-1) squared plus 1 (x less than or equal to 1)?

Is the inverse function of the function y = (x-1) squared plus 1 (x less than or equal to 1)?

y=(x-1)^2+1,x=1
The solution is 1-x = √ (Y-1)
That is, x = 1 - √ (Y-1)
So the inverse function is y = 1 - √ (x-1), x > = 1

The inverse of cosy [- x] π

As shown in the figure y = cosx x x ∈ [- π, 0], the image is a black curve
Y = arccosx x ∈ [- 1,1] the image is yellow curve
The image of the inverse function is a red curve
Y = arccos (- x) - π x ∈ [- 1,1]. Or y = - arccos x x ∈ [- 1,1]

What is the inverse function of y = x-cosx,

Can't write a concrete expression, transcendental equation

The inverse function of the function y = cosx, X ∈ (- π, - π / 2) is_______

x∈(-π,-π/2),
∴x+π∈(0,π/2),
∴cos(x+π)=-y,
x+π=arccos(-y)=π-arccosy,
∴x=-arccosy,
x. Y = - arccosx, X ∈ (- 1,0) is obtained

Inverse function of y = (cosx) ^ (3 / 2)

Use the definition to get the answer directly

Y = cosx, X belongs to [5 / 2,3] school of inverse function

Y = arccosx + 2pi x belongs to [- 1,0]

How to find the inverse function of y = a (x-cosx + 1)? My original intention is to know what shape of wavy ramp can make the velocity sine with time. That is to say, given V (T), find H (s) Note: t_ 0,v_ t,h_ m. Medium_ 0, T, M is the corner mark, = > is the launch number Constant: T_ 0,v_ 0 is unit time, unit distance, H_ M is the maximum height and G is the acceleration of gravity The ball travels on a smooth wavy slope of continuous period, set up t=(t_ 0)α V_ t=(1+sinα)v_ 0, which is V (T) When t, then: Kinetic energy E_ k=1/2(1+sinα)^2v_ 0^2m, The kinetic energy is transformed into gravitational potential energy. Let the altitude be 0 when the velocity is maximum, and the velocity is at the maximum height 0, so there are: E_ Total = e_ k+E_ h=2mV_ 0^2=mgh_ m ,=>g=2v_ 0^2/h_m > E_ h=mgh=E_ Total - e_ k=2mv_ 0^2-1/2m(i+sinα)^2v_ 0^2 > h=[4-(1+sinα)^2v_ 0 ^ 2] / 2G, which is h (α) Look at the relationship between S and α, s = integral (0, α) [(1 + sin α) v_ 0t_ 0]dα => s=v_ 0t_ 0 (α - cos α + 1), which is s (α) Now, if you want to find H (s), you need to know the inverse function α (s) of S (α), and then substitute The image of H (s) and H (s) obtained by H (α) is actually the shape of wavy ramp Y = a (x + cosx + 1) is s (α) mentioned above. It is not necessary to look at the reasons above, as long as the inverse function is given, it is better to give the solution process (correction: v_ 0 is the unit speed) You can talk about it in detail! In addition, I can prove that y = a (x-cosx + 1) is monotonically increasing, and there is no limit to the domain of the inverse function, because the derivative is y = a (SiNx + 1) > = 0 I found that y = x-cosx is a kind of translation of y = SiNx + X with function image software. If the latter can not get the inverse function, then the former should not get... Frustrated! How can any problem remain unsolved in such a developed mathematics today? Is there any other way to solve that physics problem?

The inverse function of this function is transcendental. I'm afraid there will be no definite analytic solution
As long as X and COS x, SiN x and so on appear together, there is often no analytic solution, such as x + SiN x = 1. Although this equation is simple, we can only find the approximate value of its solution, and can not express it with an analytic formula

How to find the inverse function of y = x ^ 2-2x + 2 (x less than or equal to 1)?

Y=X^2-2X+2
y=(x-1)^2+1
y-1=(x-1)^2
X-1 = ± √ Y-1 because x is less than or equal to 1,
therefore
x-1=-√y-1
That is, x = 1 - √ Y-1
therefore
The inverse function is as follows:
y=1-√x-1,x>=1

If the inverse function F-1 (x) = 1 + X2 (x < 0), then f (2) = () A. 1 B. -1 C. 1 and - 1 D. 5

According to the meaning of the title, 2 = 1 + X2 (x < 0),
The solution is x = - 1
Therefore, B is selected

Given the function y = f (x) real odd function, when x is greater than or equal to 0, f (x) = 3 x power minus one, Let f (x) inverse function be g (x) = x, then G (x) =? Wrong, the question is g (- 8) =?

Y = f (x) real odd function, when x is greater than or equal to 0, the x power of F (x) = 3 minus one
-->When x < 0, f (x) = - (3 ^ (- x) - 1) -- > 1-f (x) = 3 ^ (- x) -- > x = - log3 (1-f (x))
-->g(-8)=-log3(1-(-8))=-2