If f(x) = 1+ 1/(x-1), prove that f(x) is its own inverse?

Let's find the inverse:

Using y instead of f(x) for now:

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

y = 1 + 1/(x - 1)

Now switch the variables and solve for y again:

x = 1 + 1/(y - 1)

Multiply both sides by (y - 1):

x(y - 1) = 1(y - 1) + 1

Expand:

xy - x = y - 1 + 1The right side simplifies:

xy - x = y

Move all terms with "y" to the left and all others to the right:

xy - y = xFactor out the y:

y(x - 1) = x

y = x / (x - 1)

Now to check to see if that's the same, let's simplify the original equation:

y = 1 + 1 / (x - 1)

Common denominator:

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

Add the numerators:

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

simplify:

y = x / (x - 1)And both the inverse and the simplified original equation are the same.