MHB Function Composition: Restrictions for Domains

[Dustinsfl]

Given three functions $F,G,H$ what restrictions must be placed on their domains so that the following four composite functions can be defined?
$$
G\circ F,\quad H\circ G,\quad H\circ (G\circ F),\quad (H\circ G)\circ F.
$$

I need a hint or something.

 

[Fantini]

I'd say that in each case you'd need $\textrm{Im }F \subset D_G, \text{ I am }G \subset D_H, \text{ I am }G \circ F \subset D_H$ and $\text{Im }F \subset D_{H \circ G}$.

Taking as example your other thread, we need $\text{Im }F \subset D_G$, where $F(x) = x+5$ and

$$G(x) = \begin{cases} \frac{|x|}{x}, & \text{if } x \neq 0 \\ 1, & \text{if } x= 0. \end{cases}$$

For $\text{Im }F$ to be contained in the domain of $G$, we need to investigate the cases where $F(x) \neq 0$ and $F(x)=0$. In particular, here we have $\text{Im }F = \mathbb{R}$ and $D_G = \mathbb{R}$ since it is defined everywhere (although it isn't continuous).

 

[Sudharaka]

Given three functions $F,G,H$ what restrictions must be placed on their domains so that the following four composite functions can be defined?
$$
G\circ F,\quad H\circ G,\quad H\circ (G\circ F),\quad (H\circ G)\circ F.
$$

I need a hint or something.

Hi dwsmith, :)

If \(G\circ F\) is to be defined properly the co-domain of \(F\) should be a subset of the domain of \(G\). Therefore the only restrictions in defining the above mentioned compositions are,

\[\mbox{codom }(F)\subseteq\mbox{dom }(G)\mbox{ and }\mbox{codom }(G)\subseteq\mbox{dom }(H)\]

Kind Regards,
Sudharaka.