Pre-Calculus: Inverse Functions?

Question

I am having some problems trying to find the inverse formula for this function:

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

I was trying to use natural log and ended up here:

ln(y) = (1+x)/(1-x) ----> I am unsure of where to proceed from here.

Answer 1

You CAN'T take the logarithm of the RHS that way!

Instead, solve for e^x first:

e^x = (y - 1) / (y + 1), so that

x = ln [(y - 1) / (y + 1)].

Live long and prosper.

Answer 2

Exchange y and x in the original function definition and solve the new y in terms of the new x. Because the range (value the old y could take) of the old function was restricted to y > 1 or y < -1, the new domain is similarly restricted to x > 1 or x < -1. Because no values in this domain will result in the new y = 0, we need not further restrict the domain to eliminate this value from the range.

x = (1 + e^y) / (1 - e^y); x > 1 or x < -1
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) / (x + 1) = e^y
ln [(x - 1) / (x + 1)] = y; x < -1 or x > 1

When the derivation is done this way, the argument of the logarithm will always be positive. However, if the derivation is done such that the logarithms of x - 1 and x + 1 are taken independently, one must work around the fact that both x - 1 and x + 1 become negative when x < -1.

Answer 3

When finding the inverse function always substitute x by y and and vice versa. so we have, y = x^2 -5 exchanging the value y by x and x by y, x = y^2 -5 now you solve for y, y^2 = x+5 y = square root of ( x+5) Therefore, f inverse of x = square root of (x+5) Now, the given x ≥ 0 says that x can be only the value greater than or equal to zero that means, no negative numbers so, out domain : [ 0, + infinity) range: [ 2.236 , + infinity) I got 2.236 by subsituting x= 0 and solving on f inverse of x.

Answer 4

Exchange x and y in the equation and resolve for y:
x=(1+e^y)*(1-e^y)
Expand the RHS
x=1-e^(2y)
1-x=e^2y
Take natural log on each side to eliminate the power:
ln(1-x)=2y*ln(e)
ln(e)=1
ln(1-x)=2y
y=(1/2)*ln(1-x) is the inverse function.

Answer 5

You made a mistake in that. If you took ln of both sides you would have
ln(y) = ln((1 + e^x)/(1-e^x)) = ln(1+e^x) - ln(1-e^x)

Try it this way instead. For convenience, instead of writing e^x over and over, I'll say z=e^x:

y=(1+z)/(1-z)
y(1-z) = 1+z
y - yz - z = 1
y -z(y + 1) = 1
z = (y-1)/(y+1)
e^x = (y-1)/(y+1)
x = ln[(y-1)/(y+1)]

Answer 6

you interchange x and y and then 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)
(x-1)/(x+1) = e^y
ln [(x-1)/(x+1)] = y

Answer 7

ah i should know this .. i forgot all the math i know during the summer.. but i know that lnx is the inverse of e^x

i hope this helps;