fill in each blank so that the resulting statement is true.\nin order to graph $y = \\frac{1}{3}\\tan3x$, an…

fill in each blank so that the resulting statement is true.\nin 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 $x =$ _______.\nin 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 \nthus two consecutive asymptotes occur at $x =$ \nand $x =$

fill in each blank so that the resulting statement is true.\nin 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 $x =$ _______.\nin 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 \nthus two consecutive asymptotes occur at $x =$ \nand $x =$

Answer

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}<x<\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}<x<\frac{\pi}{6}). Two consecutive asymptotes occur at (x =-\frac{\pi}{6}) and (x=\frac{\pi}{6}).