Q: show that a function is continuous at a given interval? Show that $f(x)= 1-\sqrt{1-x^2}$ is continuous on the interval $[-1,1]$.
Show how I work with such a question please. I have watched many youtube videos, but I can't seem to get the idea.

A: Let $g(x)=1-x^2$ and
$h(x)=\sqrt{x}$.
$h$ is continuous at $\color{red}{[0,+\infty)}$.
$g$ is continuous at $[-1,1]$ and
$g([-1,1])\subset \color{red}{[0,+\infty)}$ 
since $\forall x\in[-1,1] \;\; 1-x^2\ge 0$
thus
$$ f=h(g) \text{ is continuous at } [-1,1].$$
