Maximum and minimum of two continuous functions

The formulas. For real numbers $u,v$,
$$\max(u,v)=\frac{u+v+|u-v|}2,\qquad \min(u,v)=\frac{u+v-|u-v|}2.$$
To check the first one, distinguish $u\ge v$ (then $|u-v|=u-v$ and the right side is $u$) and $u<v$ (then it is $v$). The second is similar.

Conclusion. The functions $f+g$ and $f-g$ are continuous (sums of continuous functions), so $|f-g|$ is continuous (previous problem). Hence $\max(f,g)=\frac12(f+g+|f-g|)$ and $\min(f,g)=\frac12(f+g-|f-g|)$ are continuous.

Application. Setting $f^+=\max(f,0)$ and $f^-=\max(-f,0)$ shows that the positive and negative parts of a continuous function are continuous, and $f=f^+-f^-$.