Show that $f(x) = \sqrt{1-x^2}$ is uniformly continuous on $[-1,1]$.
I showed it was differentiable on the open interval (-1,1), which implies continuous on $(-1,1)$. However, I still need to show continuity at $1$ and $-1$ so I can use the fact that continuous on a closed interval implies uniform continuity.
Note that $\sqrt x$ is continuous on $[0, \infty)$. Thus $\sqrt{1-x}$ and $\sqrt{1+x}$ are continuous on $[-1, 1]$. Hence
$$\sqrt{1-x^2} = \sqrt {1-x}\sqrt{1+x}$$
is also continuous on $[-1, 1]$.