y=f(x)=[a^x+a^(-x)]/2
y=f(-x)=[a^(-x)+a^(x)]/2=f(x)
So y = f (x) = [a x+a (-x)]/2 is an even function.
Parity function judgment rule
The domain of y=f(x) is symmetric about the origin.
(1) If f(x)+f(-x)=0, then f(x) is odd function.
(2) If f(x)=f(-x), then f(x) is an even function.