Long problem (Question 1)

Problem:



Solution:

1a) The only restriction is that the denominator cannot be 1, therefore the domain is $ (-\infty, -1) \cup (-1, +\infty) $.

1b) $ f(x) = \frac{x}{x+1} = \frac{x + 1 - 1}{x + 1} = 1 - \frac{1}{x + 1} $.
As $ x $ increase from $ -0.99.. $, the number changes from $ -\infty $ to $ 1 $.
As $ x $ decrease from $ -1.00... $, the number changes from $ 1 $ to $ +\infty $.
Therefore the domain is $ (-\infty, 1) \cup (1, \infty) $.

1c) The find the inverse function, we solve
$ y = \frac{x}{x + 1} $
$ xy + y = x $
$ y = x - xy $
$ y = x(1 - y) $
$ x = \frac{y}{1 - y} $

Therefore $ f^(y) = \frac{y}{1-y} $.

1d) Unless the domain of $ h(x) $ is a single number, otherwise it is no injective and therefore no inverse exists.