Basic composite function

Question

If $f(x)={1\over x}$ and $g(x)=\sin(x)$

Just checking if I'm understanding this correctly. Are the formulas below correct?

$$f(g(x))={1\over \sin(x)}\\ g(f(x))=\sin\left({1\over x}\right)$$

And their domains for both would be: $(x\ne0)$?

Almost all correct. For the first, there are a lot more $x$ to exclude, all integer multiples of $\pi$. — André Nicolas, Sep 04 '15 at 05:39

Vincenzo Oliva · Accepted Answer · 2015-09-04T08:45:20.367

6

Almost there. The first is not defined at integer multiples of $\pi$. If we agree on calling $D_{\phi}$ the domain of the function $\phi(x)$, then we can say $D_{f\circ g} = \mathbb{R}−\{k\pi | k\in \mathbb{Z}\}.$

edited Sep 04 '15 at 08:45

answered Sep 04 '15 at 05:40

Vincenzo Oliva

7,249

ok I'm not saying you're wrong but how so? – stuart Sep 04 '15 at 05:53
@stuart: $\sin(k\pi)=0$ for any integer $k$. Since $f(g(x))$ has this at the denominator, we need to exclude those values too from its domain. – Vincenzo Oliva Sep 04 '15 at 05:56
I see it now. Thanks. How would you use domain notation to describe this? – stuart Sep 04 '15 at 06:36
@stuart: You're welcome. I added that to my answer. :) – Vincenzo Oliva Sep 04 '15 at 09:37

score 1 · Answer 2 · answered Sep 04 '15 at 06:12

The domain for the second one is OK, because $\frac1x$ is defined for all nonzero real numbers, and $\sin \frac1x$ will then also be defined for all nonzero numbers because $\sin x$ is defined for all real $x$.

You are also correct in that $\frac{1}{\sin x}$ is not defined for $x=0$, but you should not have stopped there.

$$\frac{1}{f(x)}$$ is undefined whenever $f(x) = 0$, and $\sin x = 0$ has more than one solution. For example, because $\sin \pi = 0$, you know that $\frac{1}{\sin \pi}$ is also not defined.

Basic composite function

2 Answers2