Find a formula for the inverse of the functions.?

Question

f(x)=(10x-1)/(2x+9)

f^-1(x)= ???


y=(1+e^x)/(1-e^x)

y^-1(x)= ???

Answer 1

To find the inverse of a function, replace x with y and solve for y, or replace y with x and sove for x.

For your first question, we write

f inverse (x) = (10y - 1) / (2y + 9)

and solve for y.

x (2y + 9) = 10y - 1
2xy - 10y + 9x = -1
y (2x - 10) = -9x - 1
y = (-9x - 1) / (2x - 10)


Fixing the signs:
f inverse (x) = (9x + 1) / (10 - 2x)

To check, you would find the inverse of this function and see if you get back to the original:

x = (9y + 1) / (10 - 2y)
x (10 - 2y) = 9y + 1
10x - 2xy -9y = 1
y(-2x - 9) = -10x + 1
y = (-10x + 1) / (-2x - 9)

Fixing the signs:
f(x) = (10x - 1) / (2x + 9)

Yes!!!

For the second:

f inverse (x) = (1 + e^y) / (1 - e^y)

and solve for y:

x (1 - e^y) = 1 + e^y
x - xe^y = 1 + e^y
x - 1 = xe^y + e^y
x - 1 = e^y (x + 1)
ln ( x - 1) = y + ln (x + 1)
y = ln (x - 1) - ln (x + 1)

To check:

x = ln (y - 1) - ln (y + 1)
x = ln ((y - 1) / (y + 1))
e^x = (y - 1) / (y + 1)
ye^x + e^x - y = -1
y (e^x - 1) = -1 - e^x
y = (-1 - e^x) / (e^x - 1)

Fixing the signs:
f(x) = (1 + e^x) / (1 - e^x)

Yes!!!

Answer 2

f(x) = (10x-1)/(2x+9)
y = (10x-1)/(2x+9)...For inverse switch x and y
x = (10y-1)/(2y+9)
Solve for y
y = (9x+1)/(10-2x), so
f^-1(x) = (9x+1)/(10-2x)


y = (1+e^x)/(1-e^x)
x = (1+e^y)/(1-e^y)
Solve for y:
y = ln((x-1)/(x+1)), so
y^-1 = ln((x-1)/(x+1)) = ln(x-1) - ln(x+1)

Answer 3

you just have to switch the x fot the y and vice versa. In the inverse formula your x's become your y's