I am struggling with a question based on finding the domain of a composite function. I have managed to complete questions such as in the following document at the bottom of page three:
However, this question feels different and I am unsure of how the answer was gained:
Let $f$ and $g$ be the functions:
$$ f : [0, 2\pi] \to \mathbb R, \quad f(x) = \cos (x), $$
and
$$g : [0, 20] \to\mathbb R, \quad g(x) = x^2$$
Determine the domain for each of the compositions
$f \circ g$ and $g \circ f$. "
The given answers are:
$$\operatorname{Dom}(f \circ g) = [0,\sqrt{2\pi} ]$$
and
$$\operatorname{Dom}(g \circ f) = \:\left[0,\frac{\pi }{2}\right]\:\bigcup \:\left[\frac{3\pi }{2},\:2\pi \right]$$
Could anybody help me in working out how to find these domains please?
Thank you,
Lewis
Best Answer
Let's consider $f\circ g$; for $x$ in the still to find domain, we have $$ f\circ g(x)=f(g(x)) $$ so in particular $g(x)\in[0,2\pi]$, because otherwise we can't compute $f$ on it. On the other hand, whenever $g(x)\in[0,2\pi]$, we can compute $f(g(x))$. So we need to ensure $$ 0\le x^2\le 2\pi $$ that is, $$ x\in[-\sqrt{2\pi},\sqrt{2\pi}] $$ However, we must take into into account that $0\le x\le 20$. So we get $$ x\in[-\sqrt{2\pi},\sqrt{2\pi}]\cap[0,20]=[0,\sqrt{2\pi}] $$
For $g\circ f$ it's similar: we need $f(x)\in[0,20]$, that is $$ 0\le \cos x\le 20 $$ with the condition that $x\in[0,2\pi]$. The cosine is always at most $1$, so we just need $\cos x\ge0$, which happens for $x$ in the sets of the form $$ [0+2k\pi,\pi/2+2k\pi]\cup[3\pi/2+2k\pi,2\pi+2k\pi] $$ with $k$ integer. Intersecting with $[0,2\pi]$ gives $$ [0,\pi/2]\cup[3\pi/2,2\pi] $$