# Thread: Find inverse and prove

1. ## Find inverse and prove

I'm having trouble with two problems that I'm stuck on, I think the fact that they involve fractions is confusing me...

1) f(x) = (x+1)/x
f^-1(y) = (x+1)/x
yx = (x+1)
yx-1 = x

check:
y = (yx-1)+1/(yx-1)

and that's how far I got...

2) f(x) = 2 - 1/(x+1)
(y-2)(x+1) = -1
x+1 = -1/(y-2)
x = -1 (-1/y-2)

check:
y = 2 - 1/(-1(-1/y-2) - 1

and stuck out of my mind here

Any help would be great. Thanks

2. To find the inverse function of y = f(x), switch the variables and solve for y.

$y = \frac{x + 1}{x}$

Switch the variables.

$x = \frac{y + 1}{y}$

Now, solve for y. First, separate the fraction.

$x = \frac{y}{y} + {1}{y}$

Reduce $\frac{y}{y}$.

$x = 1 + \frac{1}{y}$

Subtract 1 from the equation.

$x - 1 = \frac{1}{y}$

Take the reciprocal of the equation.

$\frac{1}{x - 1} = y$

Therefore, the inverse function of f(x) is $y = \frac{1}{x - 1}$.

Use the same procedure for the second function.

$y = 2 - \frac{1}{x + 1}$

Switch the variables.

$x = 2 - \frac{1}{y + 1}$

Subtract 2 from the equation.

$x - 2 = -\frac{1}{y + 1}$

Reverse the sign of the equation.

$2 - x = \frac{1}{y + 1}$

Take the reciprocal of the equation.

$\frac{1}{2 - x} = y + 1$

Subtract 1 from the equation.

$\frac{1}{2 - x} - 1 = y$