Question

identify any vertical asymptotes on the interval $0,4\pi$ 17. $f(x)=2\tan(2x)$ 18. $g(x)=3\tan(\frac{x}{2})$

Explanation:

Step1: Recall the vertical asymptote formula for \(y = A\tan(Bx - C)+D\)

The vertical asymptotes of \(y=\tan x\) occur at \(x=(2n + 1)\frac{\pi}{2},n\in\mathbb{Z}\). For \(y = A\tan(Bx - C)+D\), the vertical asymptotes are given by \(Bx-C=(2n + 1)\frac{\pi}{2}\), or \(x=\frac{(2n + 1)\pi}{2B}+\frac{C}{B}\)

Step2: For \(f(x)=2\tan(2x)\)

Here \(A = 2\), \(B = 2\), \(C = 0\), \(D=0\). Using the formula \(x=\frac{(2n + 1)\pi}{2B}\), substitute \(B = 2\). Then \(x=\frac{(2n + 1)\pi}{4}\)

Find values of \(n\) such that \(x\in[0,4\pi]\)
When \(n = 0\): \(x=\frac{\pi}{4}\)
When \(n = 1\): \(x=\frac{3\pi}{4}\)
When \(n = 2\): \(x=\frac{5\pi}{4}\)
When \(n = 3\): \(x=\frac{7\pi}{4}\)
When \(n = 4\): \(x=\frac{9\pi}{4}\)
When \(n = 5\): \(x=\frac{11\pi}{4}\)
When \(n = 6\): \(x=\frac{13\pi}{4}\)
When \(n = 7\): \(x=\frac{15\pi}{4}\)

Step3: For \(g(x)=3\tan(\frac{x}{2})\)

Here \(A = 3\), \(B=\frac{1}{2}\), \(C = 0\), \(D = 0\). Using the formula \(x=\frac{(2n + 1)\pi}{2B}\), substitute \(B=\frac{1}{2}\). Then \(x=(2n + 1)\pi\)

Find values of \(n\) such that \(x\in[0,4\pi]\)
When \(n = 0\): \(x=\pi\)
When \(n = 1\): \(x = 3\pi\)

Answer:

For \(f(x)=2\tan(2x)\) on \([0,4\pi]\), the vertical asymptotes are \(x=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4},\frac{9\pi}{4},\frac{11\pi}{4},\frac{13\pi}{4},\frac{15\pi}{4}\)
For \(g(x)=3\tan(\frac{x}{2})\) on \([0,4\pi]\), the vertical asymptotes are \(x=\pi,3\pi\)