# Is the function f(x) = secx continuous on the interval [-pi/2, pi/2]?

Sep 7, 2016

No it is not.

#### Explanation:

$\sec x = \frac{1}{\cos} x$

So $\sec x$ in undefined where $\cos x = 0$, and that happens at odd multiples of $\frac{\pi}{2}$, like $- \frac{\pi}{2}$ and $\frac{\pi}{2}$.

$\sec x$ is undefined at $- \frac{\pi}{2}$ and $\frac{\pi}{2}$, so it is not continuous on the closed interval, $\left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$.

It is continuous on the open interval $\left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$.