How do I show that $\sin(x^2+3x+2)$ is not periodic? I know that if $g(x)$ is periodic then $f(g(x))$ is periodic.
This is a sufficient condition but not necessary as $\sin(x)=f(g(x))$ is periodic where $g(x)=x$(non periodic) and $f(x)=\sin(x)$.
Can the above condition be made into a necessary and sufficient condition (assuming that $g(x)$ is not the identity function) or is there a better way to find the solution?
This is from a previous year college entrance exam. Any short trick rather to check for the periodicity of $f(g(x))$ when $g(x)$ is a complicated function  would be helpful.
 A: Suppose it is periodic with period $2\pi T$ for some $T>0$.  This means that for all $x$, $\sin(x^2+3x+2)=\sin((x+2\pi T)^2+3(x+2\pi T)+2)$.  Expanding the RHS gives $\sin(x^2+3x+2)=\sin(x^2+4\pi Tx+4\pi^2T^2+3x+6\pi T+2)$.
If $\sin(u)=\sin(v)$, then either $u=v-2\pi n$ or $u=\pi-v+2\pi n$ for some constant integer $n$.  We will consider each of these cases:

*

*Suppose $x^2+3x+2=x^2+4\pi Tx+4\pi^2T^2+3x+6\pi T+2 - 2\pi n$.  Then:
$$2\pi n=4\pi Tx+4\pi^2T^2+6\pi T$$
$$n=Tx+\pi T+3T$$
$n$ is a constant, so $\frac\partial{\partial x}(Tx+\pi T+3T)=0$ (as is the case for all higher derivatives).  Calculating the derivative shows that $T=0$, but we previously defined that $T>0$, a contradiction.


*Suppose $x^2+3x+2=\pi-x^2-4\pi Tx-4\pi^2T^2-3x-6\pi T-2 + 2\pi n$.  Then:
$$2x^2+6x+4-\pi+4\pi Tx+4\pi^2T^2+6\pi T = 2\pi n$$
$$n=\frac 1\pi x^2+\frac3\pi x+\frac 2\pi-\frac 12+2 Tx+2\pi T^2+3T$$
As in the previous case, $n$ is constant, so $\frac{\partial^2}{\partial x^2}(\frac 1\pi x^2+\frac3\pi x+\frac 2\pi-\frac 12+2 Tx+2\pi T^2+3T)=0$.  Calculating this derivative gives the false statement $\frac 2\pi=0$, a contradiction.
A: A continuous periodic function is uniformly continuous.
We let $f:x\mapsto x^2+3x+2$ and we note that for $n\geq 1$, there exists unique $x_n<y_n$ such that $f(x_n) = 2n\pi$ and $f(y_n) = \pi/4+2n\pi$.
By the Mean Value Theorem, there is some $\xi_n\in (x_n,y_n)$ such that
$$\frac{\sin(f(y_n))-\sin(f(x_n))}{y_n-x_n} = (2\xi_n+3)\cos(f(\xi_n))$$
Since $f$ is increasing over $[0,\infty)$, $f(x_n)\leq f(\xi_n)\leq f(y_n)$, hence $\cos(f(\xi_n))\geq 1/\sqrt 2$ and
$$0<y_n-x_n\leq \frac{1}{2\xi_n+3}.$$
As a consequence, $y_n-x_n\to_n 0$, but $\sin(f(y_n))-\sin(f(x_n)) = 1/\sqrt 2$, hence $\sin\circ f$ is not uniformly continuous, hence not periodic.
A: We show the following lemma first:
Lemma.  A continuously differentiable differentiable periodic function $f:\mathbb{R}\to\mathbb{R}$ has bounded derivative.
Proof. Suppose $f$ has period $T$. It is clear that $f'$ must also then have period $T$. By continuity, $\lvert f'\rvert$ attains a maximum on $[0,T]$, and by periodicity
$$\max_{x\in[0,T]}\lvert f'(x)\rvert =\max_{x\in\mathbb{R}}\lvert f'(x)\rvert.$$
Thus $f'$ is bounded. $\square$
We now move on to your function
$$f:\mathbb{R}\to\mathbb{R},\quad x\mapsto\sin(x^2+3x+2).$$
We can now prove that $f$ is not periodic if we can prove that $f'$ is unbounded. Now we have that
$$f'(x)=(2x+3)\cos(x^2+3x+2).$$
But notice that then, if we choose a sequence $\{x_n\}_{n\in\mathbb{N}}$ such that $x_n\to\infty$ and $x_n^2+3x_n+2=2\pi k_n$ for some sequence of integers $\{k_n\}_{n\in\mathbb{N}}$, then
$$\lvert f'(x_n)\rvert = \lvert 2x_n+3\rvert\to\infty$$
as $n\to\infty$, and so $f'$ is unbounded. It follows that $f$ is not periodic.
