Answer to Question #217359 in Calculus for Carol

Question

The function f(x)=x2sin(x/2) is?

A. Periodic with 4π

B. Periodic with 2π

C. Periodic with 4π

D. it's not periodic

Answer 1

Every continuous periodic function on $\mathbb{R}$ must be bounded. Indeed, let $T>0$ be a non-zero period of the function $f(x)$. If $f(x)$ is continuous, it is bounded on the closed interval $[0,T]$, that is $|f(x)|\leq M$ for some $M>0$ for all $x\in [0,T]$.

For arbitrary $x\in\mathbb{R}$ let $n$ be an integer such that $n\leq x/T<n+1$, then $nT\leq x<nT+T$ and $a=x-nT\in[0,T]$. Since the function $f(x)$ is $T$-periodic, then

$f(x)=f(a+nT)=f(a)$ and hence, $|f(x)|\leq M$. Therefore, $f(x)$ must be bounded.

The function $f(x)=x^2\sin(x/2)$ is unbounded, as $|f(2\pi(n+1/2))|=4\pi^2(n+1/2)^2$. Therefore, it is not periodic.

Answer. (D) it's not periodic