Answer to Question #217359 in Calculus for Carol

Question #217359

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




1
Expert's answer
2021-08-10T15:25:42-0400

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


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