# Question #41b35

Apr 19, 2015

You know the period is $\setminus \pi$ because you know the period of $\tan \left(x\right)$ is $\pi$, and you know that if a function $f$ is periodic with period $T$, also is $\alpha f , \alpha \ne 0$ because $\alpha f \left(x\right) = \alpha f \left(x + T\right)$, this is trivial (it's just multiplication on both sides of the $=$ sign).
For the graph, you know it's zero for $x = k \pi , k \setminus \in \setminus \mathbb{Z}$ and in every zero the tangent has 4 as angular coefficient, an it is $4 f \mathmr{and} x = \frac{\pi}{4} + k \pi , - 4 f \mathmr{and} x = - \frac{\pi}{4} + k \pi$, and has a polar singularity in $x = \frac{\pi}{2} + k \pi$
graph{4*tan(x) [-10, 10, -5, 5]}

Apr 19, 2015

The period is $\pi$, because tanx repeats it self in the interval $\left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$ and would repeat in successive intervals to ($\frac{\pi}{2} , \frac{3 \pi}{2}$).. and so on to the right of the origin and similarly to the left of it.

Apr 19, 2015

I am not sure it helps, but also you can "see" the period of your function considering the coefficient of x (the number in front of it called $k$);
In this case is $k = 1$ so you have that:
$k = \frac{2 \pi}{p e r i o d}$
If $k = 1$ then
$p e r i o d = 2 \pi$