The known point P (1,2) is on the image of the function y = the following sign ax-b, and at the same time find a * B on its inverse function

By substituting (1,2) into the original function, the following results are obtained:
A-B=4
His inverse function is y = (x ^ 2 + b) / A, and then substituting (1,2) into the following equation:
1+B=2A
A=-3,B=-7
A*B=21

The inverse function of the function y = √ x squared-4 (x ≤ 2) is

Y = - x ^ 2 + 2x-5 (x < 1) x ^ 2-2x + y + 5 = 0 using the formula of root x = 1 + root (- 4-y) x > 1 is not consistent with the meaning of the topic. The inverse function of x = 1-radical (- 4-y) is y = 1-radical (- 4-x)

Function y is equal to negative x squared, plus a condition, what is the inverse function of X less than or equal to zero?

Y is equal to negative root, X is greater than or equal to zero

Inverse function of the function y = x squared x greater than or equal to 0

y=x^2,x>=0
x=y^(1/2),y>=0
Inverse function: y = x ^ (1 / 2), x > = 0

What is the inverse function of y = cosx, X ∈ [- π, 0], the answer is y = - arc cosx, X ∈ [- 1,1] how to come from and why there are negative signs

The inverse function of y = cosx, X ∈ [0, π] is y = arccosx, X ∈ [- 1,1]
∵y=cosx,x∈[-π,0]
∴-x∈[0,π],
And y = cosx = cos (- x)
∴-x=rarccosy
∴x=-arccosy
The transposition of X and Y leads to
The inverse function of y = cosx, X ∈ [- π, 0] is
y=-arc cosx,x∈[-1,1]

(x) = (x) / (1) How to solve x = (1 + y) / (1-y) from y = (x-1) / (x + 1), write the steps,

y(x+1)=x-1
yx+y=x-1
x-yx=y+1
x(1-y)=y+1
x=(y+1)/(1-y)

Function y = - cosx (0

Both are right!
y＝－cosx＝cos(π－x),π－x∈(0,π)
Therefore, π - x = arccosy, that is, x = π - arccosy
So, the inverse function is y = π - arccosx
Because arcsinx + arccosx = π / 2, the inverse function can also be written as y = π / 2 + arcsinx

The inverse function of y = cosx on [PI, bixue]

The inverse function of y = cos (x) (π ≤ x ≤ 2 π) is y = π + arccos (x), (- 1 ≤ x ≤ 1)

Y = inverse function of cosx + 1, What is the inverse function of y = cosx + 1 (- PI ≤ x ≤ 0)? The first two answers are wrong, the third one can't understand!

cosx=y-1
-PI ≤ x ≤ 0, because the arccosine is defined on 0 ≤ x ≤ PI and the cosine is an even function
So x = - acrcos (Y-1)
The inverse function is y = - arccos (x-1)
The definition domain is the original function value domain, namely (0,2)
It is a supplement to the fourth floor. It is mainly to understand the definition of the arccos function and stipulate the symbol acrcos
Directly defined on (0, PI)
For example, if a = acrcos (x), then there must be 0 ≤ a ≤ PI

Is it not true that the inverse function of y = cosx is y = arccosx? (π) Math homework for users on October 31, 2017 report Use this app to check the operation efficiently and accurately!

Right
Because y = cosx is at (π), the inverse function of y = cosx is y = arccosx

The known point P (1,2) is on the image of the function y = the following sign ax-b, and at the same time find a * B on its inverse function