Finding the domain of $f \circ g$ where $f(x) = \sqrt{x}$ and $g(x) = \frac{x}{x-1}$.

Question

If $f(x) = \sqrt{x}$ and $g(x) = \dfrac{x}{x-1}$, I want to find $D_{f \circ g}$ using the definition $$D_{f \circ g} = \{x \in D_{g}: g(x) \in D_{f} \}$$ I used that $g(x) = 1 + \dfrac{1}{x-1}$ and $D_{f}= [0, \infty[$. but I reach a contradiction that $x \in \mathbb{R} - \{1\}$ and $0 \geq x > 1$, so could anyone tell me the correct answer?

Thanks!

Answer 1

\begin{align} D_{f \circ g} &= \{x \in D_{g}: g(x) \in D_{f} \}\\ &= \{x \in \mathbb{R} - \{1\}: \dfrac{x}{x-1} \in D_{f} \}\\ &= \{x \in \mathbb{R} - \{1\}: \dfrac{x}{x-1} \geq0 \}\\ &= \{x \in \mathbb{R} - \{1\}: x\in(-\infty,0]\cup(1,\infty) \}\\ &= (-\infty,0]\cup(1,\infty) \end{align}

Answer 2

We have $x \in D_{f \circ g}$ $ \iff x>1$ or $x \le 0$.

Answer 3

$x$ being in $f\circ g$'s domain is equivalent to $$\frac{x}{x-1}\geq0$$ where the left side must be defined first of all to evaluate to be $\geq0$. This will make $x$ valid input for $g$, and then make $g(x)$ valid input for $f$. The inequality is tricky because of issues with $0$, negative numbers, and inequalities. Here is one way to solve it.

$$\begin{align} \frac{x}{x-1}&\geq0\\ 1+\frac{1}{x-1}&\geq0\\ \frac{1}{x-1}&\geq-1\\ x-1&>0&\text{ or}&&x-1\leq-1\\ x&>1&\text{ or}&&x\leq0 \end{align}$$

At the step where the split happens, either $x-1$ is positive so of course its reciprocal is greater than $-1$; or $x-1$ is negative, and taking the reciprocal of both sides reverses direction of the inequality.