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So I'm pondering what exactly this calculation entails, as it surely isn't as simple as checking the limits of the first f(t) function and deciding which of the two functions to use for the injection of the number. Formatting what I have on my hands here:

$$f(t) = t(t^2-\pi^2) -\pi \le t <\pi; f(t+2\pi)=f(t)$$

And I'm to solve

$$f(3π)$$

but surely it's something more complex than deciding that the second function is to be used and simply injecting 3pi to the t. The topic is Fourier's theorem / calculus.

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    Calculate $f(-\pi)$ using the definition of $f$ on $[-\pi,\pi)$. Then calculate $f(\pi)$ by using the relation $f(t+2\pi)=f(t)$ with $t:=-\pi$. Then calculate $f(3\pi)$ by using the relation $f(t+2\pi)=f(t)$ with $t:=\pi$. – 2017-01-24
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    Note that $f$ is $2\pi$-periodic. Hence the value of $f$ at $3\pi$ equals the value of $f$ at $3\pi-2\pi k$ for every $k\in\mathbb{Z}$. Take the $k$ such that $3\pi-2\pi k\in[-\pi,\pi)$ (the interval where you know the values of $f$). – 2017-01-24

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Note that if $t=-\pi$, then $$f(-\pi)=0 \implies f(\pi)=0 \implies f(3 \pi)=0$$ So we have that $$f(3 \pi)=0$$

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    Would this lead to f(2π) = 0 as well (Or any integer times π at that)? – 2017-01-24
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    @Grak No. $f(2 \pi)=f(0)=-\pi^2$. However, we do have that any odd number times $\pi$ is a root of $f(x)=0$. – 2017-01-24