Question

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

Answer 1

No it is not.

Explanation:

`secx = 1/cosx`

So `secx` in undefined where `cos x =0`, and that happens at odd multiples of `pi/2`, like `-pi/2` and `pi/2`.

`secx` is undefined at `-pi/2` and `pi/2`, so it is not continuous on the closed interval, `[-pi/2,pi/2]`.

It is continuous on the open interval `(-pi/2,pi/2)`.