The behavior of a periodic function repeats every period. The function is increasing on the interval from $$m=1$$ to $$m=7$$. This behavior will repeat 12 months later. The interval starting 12 months after $$m=1$$ is $$m=1+12=13$$, and the interval ending 12 months after $$m=7$$ is $$m=7+12=19$$. Therefore, the function must be increasing from $$m=13$$ to $$m=19$$.

Since the function is periodic with period 6, if a minimum occurs at $$t=2$$, then other minima must occur at $$t = 2 + 6k$$ for integer values of $$k$$. For $$k=2$$, a minimum occurs at $$t = 2 + 6(2) = 14$$. The locations of maxima and intervals of increase/decrease depend on the specific shape of the function within a period and cannot be determined from the given information alone.

For a function to be periodic, the output values must repeat exactly over successive intervals of a fixed length (the period). While the function $$f(x) = x \sin(x)$$ oscillates, the factor of $$x$$ causes the amplitude of the oscillations to increase as $$|x|$$ increases. Since the maximum and minimum values change in each oscillation, the function's values do not repeat exactly, and therefore it is not periodic. Distractor D is not a sufficient reason, as some products can be periodic.

The definition of a periodic function $$f$$ with period $$k$$ is that the function's values repeat every $$k$$ units. This is formally expressed as $$f(x+k) = f(x)$$ for all $$x$$ in the function's domain, where $$k$$ is the smallest positive constant for which this is true. Distractor C is the definition of an even function. Distractors B and D describe other types of function transformations, not periodicity.

The period of a periodic function is the length of the smallest interval over which the function completes one full cycle. The description states that the part moves from its minimum, to its maximum, and back to its minimum in 8 seconds. This is the definition of one full cycle. Therefore, the period is 8 seconds. Distractor A is the half-period (time from min to max).

Since the function is periodic with a period of 12 months, the behavior of the function on any interval will repeat 12 months later. The interval $$14 < t < 17$$ corresponds to the interval $$ (2+12) < t < (5+12) $$. Therefore, the population must also be decreasing on this interval. The other intervals do not correspond to a simple period shift of the given interval.

A periodic function with period $$k$$ satisfies the property $$f(x + k) = f(x)$$. Since the period is 12, $$f(5 + 12) = f(5)$$. Therefore, $$f(17) = f(5) = 10$$. Distractor B incorrectly assumes the function value at the period is equal to a known value. Distractor C incorrectly uses $$f(8+12) = f(20) = 3$$, not 10. Distractor D confuses periodicity with properties of odd functions.

The time between a consecutive maximum (high tide) and minimum (low tide) represents half of one full cycle. Therefore, the period is twice this duration. The period is $$2 \times 6.2 = 12.4$$ hours. Distractor B is the half-period. Distractor A is a quarter-period. Distractor D is two full periods.

The period is the length of one cycle. If 4 cycles are completed over an interval of length 20, the length of one cycle is the total length divided by the number of cycles. Period = $$\20 / 4 = 5$$. Distractor A is the number of cycles. Distractor D is the length of the entire interval shown. Distractor C is the length of two cycles.

The function is periodic with a period of 16 milliseconds. If a maximum occurs at $$t=4$$, subsequent maxima will occur at $$t = 4 + 16k$$ for any integer $$k$$. For $$k=1$$, the next maximum is at $$t = 4 + 16 = 20$$ milliseconds. Other options do not follow this pattern.