Question

max{f(x),g(x)}=1/2(f(x)+g(x)+|f(x)-g(x)|

Answer 1

f(x)≥g(x)时,max{f(x),g(x)}=f(x)=1/2(f(x)+g(x)+f(x)-g(x))=1/2(f(x)+g(x)+|f(x)-g(x)|).

f(x)＜g(x)时,max{f(x),g(x)}=g(x)=1/2(f(x)+g(x)+g(x)-f(x))=1/2(f(x)+g(x)+|f(x)-g(x)|).

所以,max{f(x),g(x)}=1/2(f(x)+g(x)+|f(x)-g(x)|).