## Calculus 10th Edition

$f(x)$ is continuous over the following infinite number of intervals: $...(-10, -6)$U$(-6, -2)$U$(-2, 2)$U$(2, 6)$U$(6, 10)...$
Using Theorem $1.12$ with $f(x)=h(g(x));h(x)=\sec{x}$ and $g(x)=\dfrac{\pi x}{4}:$ $g(x)$ is continuous for any value of $x$ but $h(x)$ is continuous for $x\ne\dfrac{\pi}{2}+k\pi$ where $k$ is any integer. $g(x)\ne\dfrac{\pi}{2}+k\pi\to\dfrac{\pi x}{4}\ne\dfrac{\pi}{2}+k\pi\to x\ne2+4k$ $h(g(x))=f(x)$ is continuous for all values of $x$ such that $x\ne2+4k.$ By plugging in some values for $k$, we get the intervals to be: $...(-10, -6)$U$(-6, -2)$U$(-2, 2)$U$(2, 6)$U$(6, 10)...$