Find the fundamental period of the function $\large f(x) = e^{3(x - \lfloor x \rfloor )}$ .

**
Notation:
**
$\lfloor \cdot \rfloor$
denotes the
floor function
.

The answer is 1.

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Since $x+1-\lfloor x+1 \rfloor = x-\lfloor x \rfloor$

But $x+\frac{1}{2}-\lfloor x+\frac{1}{2} \rfloor \neq x-\lfloor x \rfloor$

The period of the function is $1$ .