Question

What is the period of the graph tan(3pix +4)?

Answer 1

2/3

Answer 2

Recall that `f(kx)` shrinks the graph of `f(x)` horizontally by a factor of `k`.

If you knew that `f(x)` is periodic with a period of `P`,
then you would also know the period of the graph shrunk horizontally by a factor of `k` :
The period of `f(kx)` is simply `P/k`.

Answer 3

We know the period of tan is pi since it's constructed from:

\[\frac{\sin(x)}{\cos(x)} = \frac{\sin(x+\pi)}{\cos(x+\pi)}=\frac{-\sin(x)}{-\cos(x)} =\frac{\sin(x)}{\cos(x)} \]

Now we know phase shift doesn't affect the period, so we can remove it so it's not a distraction and look at our period and compare:
\[\tan(3\pi x )=\tan(3\pi (x+T) ) = \tan(3\pi x + 3\pi T)\]
We know however that:
\[3\pi T = \pi\] since it's the period of tan. So we solve for T now which is just 1/3.

Answer 4

To me,
\[tan(3\pi x +4) =\dfrac{sin(3\pi x+4)}{cos(3\pi x +4)}\]

Hence \(tan (3\pi x +4) =0 \longleftrightarrow sin(3\pi x +4) =0 \)

Then we solve for value of x where tangent =0. At the first point, I let sin 0 =0 to solve for x, then I got \(x_1= -\dfrac{4}{3\pi}\)

then let \(sin \pi =0\) and solve for x , I got
\(x_2 = \dfrac{1}{3}-\dfrac{4}{3\pi}\)

the difference between \(x_1,x_2\) is the period. Hence the period is 1/3