Answer:
[tex]x\neq 1+2n,n\in\mathbb{Z}[/tex]
Step-by-step explanation:
Remember that for the tangent parent function, it has infinite discontinuities (vertical asymptotes) on:
[tex]f(x)=\tan(x), \\\text{Where }x\neq\frac{\pi}{2}+n\pi, n\in\mathbb{Z}[/tex]
Here, we have:
[tex]f(x)=\tan(\frac{\pi x}{2})[/tex]
So, we can set the expression inside the tangent to equal our parent domain restriction. This yields:
[tex]\frac{\pi x}{2}\neq\frac{\pi}{2}+n\pi[/tex]
Solve for x. Multiply both sides by 2:
[tex]\pi x\neq \pi+2n\pi[/tex]
Divide both sides by π:
[tex]x\neq 1+2n, n\in\mathbb{Z}[/tex]
Therefore, for the function [tex]f(x)=\tan(\frac{\pi x}{2})[/tex], it is not continuous for all Xs [tex]1+2n[/tex] where [tex]n\in\mathbb{Z}[/tex]
