Super simple inverse function, online evaluation y=2x+6 It's about the process By the way, what are even and odd functions

Super simple inverse function, online evaluation y=2x+6 It's about the process By the way, what are even and odd functions

Y = 2x + 6 x = (y-6) / 2 this is the inverse function
Even function is f (- x) = f (x), and its image is symmetric about y axis
The odd function is f (- x) = - f (x), whose image is symmetric about the origin

Find a simple inverse function Y=（X+1）/（2X-1）

y(2x-1)=x+1
2xy-y=x+1
(2y-1)x=1+y
x=(1+y)/(2y-1)
f'(x)=(1+x)/(2x-1)
In addition, y = (x + 1) / (2x-1) = (x-1 / 2 + 3 / 2) / (2x-1) = 1 / 2 + (3 / 2) / (2x-1)
That is, y ≠ 1 / 2
So the definition domain of inverse function is x ≠ 1 / 2

Find the inverse function of the function f (x) = LG (x + radical (x ^ 2 + 1))

Let y = LG (x + radical (x ^ 2 + 1)), then x + radical (x ^ 2 + 1) = 10 ^ y, so the root sign (x ^ 2 + 1) = 10 ^ y-x. the square of both sides leads to x ^ 2 + 1 = x ^ 2-2 * 10 ^ y * x + 10 ^ (2Y), so 2 * 10 ^ y * x = 10 ^ (2Y) - 1, so x = [10 ^ (2Y) - 1] / (2 * 10 ^ y) interchanges x, y into the inverse function y = 1 / 2 [10 ^ X-10 ^ (- x)]

How to find the inverse function of x-radical (x-1)

Y = root x-radical (x-1)
1 / y = 1 / (root X - root (x-1))
=Root x + root (x-1)
Add up
Y + 1 root
(y + 1 / y) / 2 = radical x
X = (Y+1/Y)^2/4.

Function y = under radical (x ^ 2 + 1) + 1 (x

y=√（x^2+1）+1>2
=> (y-1)^2 =x^2+1
=>x^2=(y-1)^2-1
=>x=-√[(y-1)^2-1]
=>x=-√[(y-1)^2-1]
=>y=-√[(x-1)^2-1] (x>2)

If the inverse function of the function y = radical (x-1) + 1 (x > = 1) is

The square of the two sides is (x-1)
So x = (Y-1) squared + 1
Because x > = 1, y = radical (x-1) + 1 > = 1
The inverse function is y = (x-1) square + 1, x > = 1

Y = radical X-1 / inverse function of radical x

y=√x-1/√x,(x>0)
y√x=x-1
x-y(√x)-1=0
Let √ x = t
Then T 2 - YT-1 = 0
t=[y+√y²+4）]/2
Or T = [y - √ (y 2 + 4)] / 2
Radical x = [y + √ (y 2 + 4)] / 2
x=[y+√（y²+4）]²/4
That is, the inverse function y = [x + √ (x 2 + 4)] 2 / 4
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What is the symmetry of the graph of the function y = f (x) and its inverse function y = f to the negative first power (x)

On y = x symmetry
Both the original function and its inverse function are symmetric about the line y = X

High school mathematics function y = f (x) (- 13) y = F-1 power (x) is its inverse function, then the image of F-1 power (x + 2) must pass through the point If the function y = f (x) (- 13) y = F-1 power (x) is its inverse function, then the image of power F-1 (x + 2) must cross the point A -5 1 B -3 3 C -3 1 D -5 3

No answer, should be too much (1, - 1)

The inverse function of the function y = e to the x power / 1 + (e to the x power) How does Y + y * e introduce Y / (1-y)

X power of Y + y * e = x power of e
The x power of E = Y / (1-y)
x=ln[y/(1-y)]
True number y / (1-y) > 0
y(y-1)<0
So the inverse function y = ln [x / (1-x)], where 0
Homework help users 2017-09-25
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