Studying the monotonicity of $g(x) = \frac{x}{x-1}$ (without using the derivative)

Question

What are the cases in which we divide the domain of the function when studying monotonicity, I know that when the function contains an absolute value, a squared quantity, but shall I divide the domain in our case, Studying the monotonicity of $g(x) = \frac{x}{x-1}$, and to which intervals shall I divide it?

Answer 1

You want to split the domain into intervals over which the function is continuous throughout. Since there is a singularity at $x=1$, The real line is split into $(-\infty,1)$ and $(1,\infty)$

Answer 2

Hopefully the following distribution can help you visualize:

$$g(x) = \frac{x}{x-1}=\frac{x-1+1}{x-1}=1+\frac{1}{x-1}$$

You might want to consider $x<1$ and $x>1$.

Answer 3

We notice that at $x = 1$ the function is discontinuous, so we'll have to split for $x < 1$ and $x > 1$. Now, you want to check during which intervals your function is increasing and when it's decreasing. So we can take a look at the derivative

$$g'(x) = \frac{-1}{(x-1)^2}$$

We see that, once again, the only interesting point is $x = 1$, because nothing else causes $g'(x)$ to be either zero or undefined. Notice that the numerator is always negative and the denominator is always positive, for all $x \neq 1$. What does this tell you?

Answer 4

For $g(x)=x/(x-1)=1-1/(x-1)$ we have

(I). For $x>1,$ the function $x-1$ is positive and increasing

so the function $1/(x-1)$ is positive and decreasing

so the function $-1/(x-1)$ is negative and increasing

so the function $-1/(x-1)+1=g(x)$ is increasing.

(II). Apply a similar method for $x<1.$