# Where are the critical points of tan x?

Jun 14, 2015

$x = \frac{\pi}{2} + k \pi \text{ where " k in ZZ}$.

#### Explanation:

If you write $y = \tan x = \sin \frac{x}{\cos} x$, when $\cos x = 0$, you have a null denominator.
The points of discontinuity of the function $y = \tan x$ are in $x = \frac{\pi}{2} + k \pi \text{ where " k in ZZ}$, that are the solutions of the equation $\cos x = 0$.
Those points correspond to a set of vertical asymptotes for the function $y = \tan x$.

graph{tanx [-10, 10, -5, 5]}

Jun 16, 2015

In the sense of critical points from calculus, which are points in the domain where the tangent line is either horizontal, doesn't exist, or has infinite (undefined) slope (if it's vertical), the function $y = \tan \left(x\right)$ has no critical points.
You can see from the graph already shown in the other answer that the function $y = \tan \left(x\right)$ never has a horizontal or vertical tangent line.
Tangent lines to $y = \tan \left(x\right)$ don't exist at $x = \frac{\pi}{2} + n \pi$ for $n = 0 , \setminus \pm 1 , \setminus \pm 2 , \setminus \pm 3 , \setminus \ldots$, however, those are also not in the domain of $y = \tan \left(x\right)$, so they technically don't count as critical points.