The inverse function of the function y = lgx (x > 0) is

The inverse function of the function y = lgx (x > 0) is

0

0

Because x ∈ R, 1 + x 2 > x 2, √ (1 + x ^ 2) > x,
X + √ (1 + x ^ 2) > 0, so y ∈ R
According to the definition of logarithmic function, 10 ^ y = x + √ (1 + x ^ 2)
10 ^ y - x = √ (1 + x ^ 2)
X = 1 / 2 [10 ^ Y-10 ^ (- y)]
y=1/2[10^x-10^(-x)] (x∈R).

If the inverse function of y = (4-x) passes through a fixed point, the coordinates of the fixed point are obtained

Y = f (x) over (0,1)
Y = F-1 (x) over (1,0)
f-1(1)=0
f-1(y)=4-x
y=1,4-x=0
Y = f (4-x) over (4,1)
Y = F-1 (x) over (1,4)

Let the inverse function of the function y = f (x) be the image crossing point P (- 2,4) of y = f ^ - 1 (x), y = f (x + 1), then the image of y = f ^ - 1 (x + 1) must pass through the point

（3,-1）
The image crossing point P (- 2,4) of y = f (x + 1)
Since y = f ^ - 1 (x) is the inverse function of y = f (x), it must go through (4, - 1) points;
Then y = f ^ - 1 (x + 1) must pass through the point (3, - 1)

If the image of the inverse function of the function y = a ^ x passes through the point (4, - 2), then a=

The image of the function y = a ^ x passes through the point (- 2,4)
By substituting the coordinates of the point into the analytic formula of the function, we can get the following results:
a^(-2)=4=(1/2)^(-2)
The solution is: a = 1 / 2
After confirmation, the above statement is correct

How to find the inverse function of y = 2x-5 / X-1 (x belongs to R and X ≠ 1)

Denominator, y = [2 (x-1) - 3] / (x-1)
y=2-[3/(x-1)]
3/(x-1)=2-y
x-1= 3/2-y
x=3/2-y + 1
x=(5-y)/(2-y)

Inverse function of y = 4x-2?

Inverse function method: invert the independent variable and dependent variable, and then invert the independent variable to represent the dependent variable
y=4x-1/2
x=4y-1/2
4y=x+1/2
y=x/4+1/8

What is the inverse function of y = 2x-4x-1 on X less than or equal to 1? Hope to have a detailed explanation

Y = 2x ^ 2-4x-1y = 2 (x-1) ^ 2-3 (y + 3) / 2 = (x-1) ^ 2 because x is less than or equal to 1, X-1 is less than or equal to 0. Therefore [(y + 3) / 2] ^ (0.5) = - (x-1) (the left is radical, greater than 0) x = 1 - [(y + 3) / 2] ^ (0.5) is the substitution of XY, and the inverse function is y = 1 - [(x + 3) / 2] ^ (0.5) ^, denotes the power, and ^ 0.5 is equal to the open root sign

If f (x) = 2x ^ 2 + 4x-7, X belongs to [0, positive infinity), then (inverse function) F-1 (- 7)=

(inverse function) F-1 (- 7) gives 2x ^ 2 + 4x-7 = - 7
X = 0 - 2 (round)
To sum up, x = 0

[mathematics of senior one] f (x) = 2x? - 4x + 1, X ∈ [- 4,0], find the inverse function

Y = 2 (x-1) ^ 2-1, when x ∈ [- 4,0], y is adjusted and subtracted, and the range is [1,49]
(x-1) ^ 2 = (y + 1) / 2
Because x ∈ [- 4,0], the formula is: 1-x = √ [(y + 1) / 2]
x=1-√[(y+1)/2]
Therefore, the inverse function is y = 1 - √ [(x + 1) / 2], X ∈ [1,49]