# The function f is periodic. If f(3) = -3, f(5) = 0, f(7) = 3, and the period of the function of f is 6, then how do you find f(135)?

Dec 9, 2015

$f \left(135\right) = f \left(3\right) = - 3$

#### Explanation:

If the period is $6$, it means that the function repeats its values every $6$ units.

So, $f \left(135\right) = f \left(135 - 6\right)$, because these two values differ for a period. By doing so, you can go back until you find a known value.

So, for example, $120$ is $20$ periods, and so by cycling $20$ times backwards we have that

$f \left(135\right) = f \left(135 - 120\right) = f \left(15\right)$

Go back a couple of periods again (which means $12$ units) to have

$f \left(15\right) = f \left(15 - 12\right) = f \left(3\right)$, which is the known value $- 3$

In fact, going all the way up, you have

$f \left(3\right) = - 3$ as a known value

$f \left(3\right) = f \left(3 + 6\right)$ because $6$ is the period.

Iterating this last point, you have that

$f \left(3\right) = f \left(3 + 6\right) = f \left(3 + 6 + 6\right) = f \left(3 + 6 + 6 + 6\right) = \ldots = f \left(3 + 132\right) = f \left(135\right)$, since $132 = 6 \cdot 22$