y=cosx（-pai/2

y=cosx（-pai/2

A: y = cosx (0

Y = arccosx, (x ∈ [- 1,1]) is the inverse function of y = cosx, (x ∈ [0, π (beat)] Correct judgment?

Of y = arccosx
The inverse function is y = cosx [π, 2 π]
The original x is from - 1 to 1, and the X of the inverse function is from π to 2 π,

Inverse function of y = x-2x (x ≤ 1)

y=x²-2x,
x²-2x-y=0
X = 1 - √ (1 + y) or x = 1 + √ (1 + y)
So the inverse function is y = 1 - √ (1 + x), X ≥ - 1

Find the inverse function of the function y = - x ^ 2 + 2x + 2 (x is less than or equal to 1)

From y = - x ^ 2 + 2x + 2
(x-1)²=3-y
∵x≤1
∴1-x=√(3-y)
∴x=1-√(3-y)
∴y=1-√(3-x)
And y = - x ^ 2 + 2x + 2 = - (x-1) 2 + 3 ≤ 3
So the definition domain of inverse function is ≤ 3
To sum up, y = 1 - √ (3-x) (x ≤ 3)

What is the inverse function of y equal to x + 2 / 2x + 1

y=（x+2)/2(x+2)-3
y[2(x+2)-3]=x+2
y[2-3/(x+2)]=1
The inverse function is
1/x=2-3/(y+2)
To be continued

What is the negative first power (x) of the inverse function f of the function y = log (x-1) (2 ≤ x ≤ 5)?

y=log(x-1)(2

If the function y = f (x) has an inverse function and f (3) = 0, then the image of function F-1 (x + 1) must pass through the point () A. （2，0） B. （0，2） C. （3，-1） D. （-1，3）

Analysis: ∵ the function y = f (x) has an inverse function and f (3) = 0,
Then the graph of function f (x) passes through point (3,0),
The graph of the inverse function F-1 (x) of the function f (x) passes through point a (0, 3),
Then F-1 (0) = 3,
Then the graph of function F-1 (x + 1) must cross the point (- 1,3)
Therefore, D

Let f (x) = (x-1 / x + 1) * (x is greater than or equal to 1). Find the - 1 power (x) of the inverse function f of F (x) The asterisk is square

sqrt(y)=x-1/x+1=1-2/(x+1)
2/(x+1)=1-sqrt(y)
x=2/(1-sqrty)-1=(1+sqrty)/(1-sqrty)
Inverse function (sqrtx + 1) / (1-sqrtx)
Sqrt is a radical

If f (x) is the sending function on R, and f (x) is equal to the x power of 1 / 2, then the value of the inverse function f (- 2) of F (x) is the value?

Because f (x) = (1 / 2) ^ x
So f (- x) = - f (x) = - (1 / 2) ^ x
Because f (x) = (1 / 2) ^ x > 0
So let f (- x) = - 2, that is - (1 / 2) ^ x = - 2
The solution is x = - 1
So the value of the inverse function f (- 2) of F (x) is - 1

Given that the function y is equal to the root sign (5-2x) × (5 + 2x) (x ∈ [0,2 / 5]), find its inverse function

If we calculate the square of both sides, X is equal to the root of half sign 25-y ^ 2. The original value range of the function is [0,5], so now the definition field of this inverse function is [0,5]