What is the domain of this composite function?

Question

The question is:

$f(x) = \dfrac{x}{x-1}$

$g(x) = \dfrac{1}{x}$

$h(x) = x^2 - 1$

Find $f \circ g \circ h$ and state its domain.

The answer the textbook states is that the domain is all real values of $x$, except $\pm 1$ and $\pm \sqrt{2}$.

However surely the domain excludes $0$ as well, since $g(0)$ is undefined.

Answer 1

You're not inputting $x$ into $g$, though. You're inputting $h(x)$. So yes, $g(0)$ is undefined, which means that whatever values of $x$ makes $h(x) = 0$ is not part of the domain. That's why they exclude $\pm1$.

Answer 2

$h(x) = x^2-1$ has a value for all $R$

$g(h(x)) = \frac1{x^2-1}$ which does not have a value at $x = \pm \sqrt1$

$f(g(h(x))) = \frac{\frac1{x^2-1}}{\frac1{x^2-1}-1}= \frac{1}{1-x^2+1}=\frac{1}{2-x^2}$ $\quad$ which does not have a value at $x=\pm\sqrt2$

Hence the total domain is $x\in R-\{\pm\sqrt1,\pm\sqrt2\}$

Answer 3

We have

• $g(h(x))=\frac1{x^2-1}$
• $f(g(h(x)))=\frac{\frac1{x^2-1}}{\frac1{x^2-1}-1}=\frac{x^2-1}{(x^2-1)(2-x^2)}$

with

• $x^2\neq1$
• $2-x^2\neq 0\implies x\neq \pm \sqrt 2$

then

• $f(g(h(x)))=\frac{1}{2-x^2}$

Answer 4

You want to find $f \circ g \circ h = f(g(h(x))) = f(g(x^2-1)) = f\bigl( \frac{1}{x^2 - 1} \bigr) = \frac{ \bigl( \frac{1}{x^2 - 1} \bigr) }{\bigl( \frac{1}{x^2 - 1} - 1 \bigr) }$

Simplifying:

$$\begin{align} f \circ g \circ h & = \frac{ \bigl( \frac{1}{x^2 - 1} \bigr) }{\bigl( \frac{1}{x^2 - 1} - 1 \bigr) } \\ & = \frac{ \bigl( \frac{1}{x^2 - 1} \bigr) }{\bigl( \frac{1 - (x^2 - 1)}{x^2 - 1} \bigr) } \\ & = \frac{ \bigl( \frac{1}{x^2 - 1} \bigr) }{\bigl( \frac{2 - x^2}{x^2 - 1} \bigr) } \\ & = \frac{x^2 - 1}{(x^2 - 1)(2 - x^2)} \end{align}$$

Here we can conclude that $x^2 - 1 \not = 0$ and $ 2 - x^2 \not = 0$

$$ \Rightarrow x \not \in \{-\sqrt{2}, \; -1, \; 1, \; \sqrt{2} \}$$

So we conclude that

$$ x \in \mathbb{R} \backslash \{-\sqrt{2}, \; -1, \; 1, \; \sqrt{2} \}$$

And you're done.

The domain of $g(h(x))$ excludes $h(x) = 0$, but this does not imply that $x=0$.