Domain of a composite function $g(h(x))$

Question

Say we have $g(x)=x-5$ and $h(x)=x^2+4x+6$. Say we have to find the composite function $g(h(x))$ and then find the domain of that function. I know that $g(h(x))=x^2+4x+1$. Now, if I had to consider the function $x^2+4x+1$ on its own and not as a composite of two other function then I can easily say that its domain is $(-\infty, \infty)$. However, I'm not sure of the domain of the composite function $g(h(x))$ would be the same since $h(x)$ has range $[2, \infty)$ and the values of $h(x)$ are the inputs of the composite function.

Answer 1

The domain is indeed $\Bbb R$. You can still put any value of $x\in(-\infty,\infty)$ into $g(h(x))$. The fact that $h$ has range $[2,\infty)$ will simply only affect the range of $g\circ h$. The range of $h$ does not affect the domain of $g\circ h$ in this case because $g$ has domain $\Bbb R$.

If $g$ were a function with domain $[3,\infty)$ for example, then the domain of $g\circ h$ would need to change, because $h(x)\in [2,\infty)$ if $x\in(-\infty,\infty)$. For instance, if $x=-2$ then $h(x)=2$, and then $g(h(x))=g(2)$ is undefined.

In general, if $f:X\subseteq \Bbb R\to Y\subseteq \Bbb R$ and $g:U\subseteq \Bbb R\to\Bbb R$, then we can compose $g\circ f$ if $Y\subseteq U$.

Answer 2

You can find the domain of any composite function using the following formula $$ D_{f\circ g} = \left\{\begin{aligned} &x\in D_g\\ &g(x)\in D_f \end{aligned} \right. $$ In your example $ g(x)=x+5,D_g=\mathbb R $ and $ h(x)=x^2+4x+6,D_h=\mathbb R $ then the domain of the composite function will be $$ D_{g\circ h} = \left\{\begin{aligned} &x\in D_h\\ &h(x)\in D_g \end{aligned} \right. = \left\{\begin{aligned} &x\in \mathbb R\\ &x^2+4x+6\in \mathbb R \end{aligned} \right. = \mathbb R $$ so $ (g\circ h)(x)=g(h(x))=x^2+4x+1 $