graph and find symmetry in this equation $y = \tan\left(\frac{\pi}{x}\right)$ I need the graph and find symmetry in this equation :
$$ y =  \tan\left(\frac{\pi}{x}\right)   $$
Could you show me the steps ?
 A: We know that $\tan(\alpha)=0$ happens when $\alpha=k\pi$, where $k$ is any integer. If, then, you set $\alpha=\pi/x$ (as in your case), we see that
$$\tan\left(\frac{\pi}{x}\right)=0 \implies x=\frac{1}{k}, k\in\mathbb{Z},k\ne0$$
That's clue number one.
For clue number two, we can find the critical points of the graph (points where the slope of the graph is horizontal). Well  $\dfrac{d}{dt}\tan(\alpha)=\sec^2(\alpha)\cdot\alpha '$, so if we once again set $\alpha=\pi/x$, we see that
$$\dfrac{d}{dt}\left(\tan\left(\frac{\pi}{x}\right)\right) = \sec^2\left(\frac{\pi}{x}\right)\cdot\left(-\frac{\pi}{x^2}\right)$$
Setting this equal to $0$, we can find the critical points.
$$\sec^2\left(\frac{\pi}{x}\right)\cdot\left(-\frac{\pi}{x^2}\right)=\frac{-\pi}{x^2\cos^2\left(\frac{\pi}{x}\right)}$$
Oh no! It looks like there are technically no critical points, but if this above expression tends to $0$ as $x$ tends to a certain value, we can say that it approaches a 'critical point'. Another way of saying this is that the denominator must tend to infinity - but if you look at the denominator ($x^2\cos^2\left(\frac{\pi}{x}\right)$), that will only tend to infinity if $x$ tends to infinity.
If, however, the denominator tends to $0$, we'll be getting closer to a vertical asymptote (slope approaches positive or negative infinity). This one is tricky, because you must not forget the $-\pi$ in the numerator. Now taking that into, consideration, we can certainly say that the denominator will be equal to $0$ at $x=\dfrac{2}{1+2k}$, where $k$ is an integer. Why? Because if you plug it into $\cos^2\left(\dfrac{\pi}{x}\right)$, you'll get zero. Additionally, the slope will always be negative! This is because $x^2$ and $\cos^2\left(\dfrac{\pi}{x}\right)$ are always positive, so the $-\pi$ term in the numerator will always win out.
And that's the second clue.
With all that said, we can make an attempt at the graph for the function.

Not so bad, eh?
