Question

fill in each blank so that the resulting statement is true.
in order to graph $y = \frac{1}{3}\tan3x$, an interval containing one period is found by solving $-\frac{\pi}{2}<3x<\frac{\pi}{2}$. an interval containing one period is _____. thus two consecutive asymptotes occur at $x =$ ___ and $x =$ _____.
in order to graph $y = \frac{1}{3}\tan3x$, an interval containing one period is found by solving $-\frac{\pi}{2}<3x<\frac{\pi}{2}$. an interval containing one period is
thus two consecutive asymptotes occur at $x =$
and $x =$

Explanation:

Step1: Solve the inequality for \(x\)

Given \(-\frac{\pi}{2}<3x <\frac{\pi}{2}\), divide each part of the inequality by 3. We get \(-\frac{\pi}{6}

Step2: Find the asymptotes

For the tangent - function \(y = A\tan(Bx - C)+D\), the vertical asymptotes occur at \(Bx - C=-\frac{\pi}{2}+k\pi\) and \(Bx - C=\frac{\pi}{2}+k\pi\), \(k\in\mathbb{Z}\). In the function \(y=\frac{1}{3}\tan(3x)\), \(B = 3\), \(C = 0\). The vertical asymptotes are found by solving \(3x=-\frac{\pi}{2}\) and \(3x=\frac{\pi}{2}\), so \(x =-\frac{\pi}{6}\) and \(x=\frac{\pi}{6}\).

Answer:

An interval containing one period is \(-\frac{\pi}{6}