判断函数y=lg(x+√ (x^2+1))的奇偶性

求详解!

y=f(x)=lg(x+√ (x^2+1))

f(-x)=lg[-x+√ (x^2+1)]

=lg[√ (x^2+1))-x]

=lg{1/[√ (x^2+1))+x]}

=lg[√ (x^2+1))+x]^(-1)

=-lg[√ (x^2+1))+x]

=-lg[x+√ (x^2+1)]

=-f(x)

∴函数y=lg(x+√ (x^2+1))为奇函数.