show that a function is continuous at a given interval?

Question

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.

Answer 1

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 $$ f=h(g) \text{ is continuous at } [-1,1].$$