what are the vertical asymptotes of the function $f(x) = \\tan\\left(\\frac{1}{2}x - \\frac{\\pi}{4}\\right)$…

what are the vertical asymptotes of the function $f(x) = \\tan\\left(\\frac{1}{2}x - \\frac{\\pi}{4}\\right)$?\n\\bigcirc a. $x = \\frac{\\pi}{2} + 2n\\pi$\n\\bigcirc b. $x = \\frac{\\pi}{6} + n\\pi$\n\\bigcirc c. $x = \\frac{\\pi}{4} + n\\pi$\n\\bigcirc d. $x = \\frac{3\\pi}{2} + 2n\\pi$

what are the vertical asymptotes of the function $f(x) = \\tan\\left(\\frac{1}{2}x - \\frac{\\pi}{4}\\right)$?\n\\bigcirc a. $x = \\frac{\\pi}{2} + 2n\\pi$\n\\bigcirc b. $x = \\frac{\\pi}{6} + n\\pi$\n\\bigcirc c. $x = \\frac{\\pi}{4} + n\\pi$\n\\bigcirc d. $x = \\frac{3\\pi}{2} + 2n\\pi$

Answer

Explanation:

Step1: Recall vertical asymptotes of tan(x)

The function ( y = \tan(u) ) has vertical asymptotes where ( u=\frac{\pi}{2}+n\pi ), ( n\in\mathbb{Z} ).

Step2: Substitute ( u = \frac{1}{2}x-\frac{\pi}{4} )

Set ( \frac{1}{2}x - \frac{\pi}{4}=\frac{\pi}{2}+n\pi ), ( n\in\mathbb{Z} ).

Step3: Solve for x

First, add ( \frac{\pi}{4} ) to both sides: ( \frac{1}{2}x=\frac{\pi}{2}+\frac{\pi}{4}+n\pi ) Simplify the right - hand side: ( \frac{\pi}{2}+\frac{\pi}{4}=\frac{2\pi + \pi}{4}=\frac{3\pi}{4} ), so ( \frac{1}{2}x=\frac{3\pi}{4}+n\pi ) Then multiply both sides by 2: ( x=\frac{3\pi}{2}+2n\pi ), ( n\in\mathbb{Z} )

Answer:

d. ( x=\frac{3\pi}{2}+2n\pi )