Answer:
![x\\eq 1+2n,n\in\mathbb{Z}](https://img.qammunity.org/2021/formulas/mathematics/college/e3m175wuj0rn0rb9us3o3e9inid6z8kcis.png)
Explanation:
Remember that for the tangent parent function, it has infinite discontinuities (vertical asymptotes) on:
![f(x)=\tan(x), \\\text{Where }x\\eq(\pi)/(2)+n\pi, n\in\mathbb{Z}](https://img.qammunity.org/2021/formulas/mathematics/college/3vyok369xxqlycxkapa4r19kojn32xpwhu.png)
Here, we have:
![f(x)=\tan((\pi x)/(2))](https://img.qammunity.org/2021/formulas/mathematics/college/w5a27bauv5w7mo7weg2gn1kavhu2zwcs51.png)
So, we can set the expression inside the tangent to equal our parent domain restriction. This yields:
![(\pi x)/(2)\\eq(\pi)/(2)+n\pi](https://img.qammunity.org/2021/formulas/mathematics/college/tqrn7ghzxvlm8agw0wdr8c9mvo4uqe280x.png)
Solve for x. Multiply both sides by 2:
![\pi x\\eq \pi+2n\pi](https://img.qammunity.org/2021/formulas/mathematics/college/8az3dhkp6hisw5v76mp29jwwhaaamsxw6e.png)
Divide both sides by π:
![x\\eq 1+2n, n\in\mathbb{Z}](https://img.qammunity.org/2021/formulas/mathematics/college/marvp0b39vdibnl203bchewuuasjh6ywsm.png)
Therefore, for the function
, it is not continuous for all Xs
where
![n\in\mathbb{Z}](https://img.qammunity.org/2021/formulas/mathematics/college/p10o5tc4uzp3x1hr1drycuymiyqy06926z.png)