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Prove that if $f(x)$ is a periodic function with period T, then the function $f(ax+b)$, where $a>0$, is periodic with period $\frac{T}{a}$.

I started with, $$f[(a(x+T/a)+b]=f[(ax+b)+T]=f(ax+b).$$ And hence $f(ax+b)$ is a periodic function. But I am not able to proceed further.

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The second part is to show that no smaller number than $T/a$ would work. The usual approach is by contradiction: assume that $k < T/a$ also works, i.e. that $f(a(x+k)+b)=f(ax+b)$ and show that this would give a smaller value for the original period than $T$.

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