The function f(x)=x+1x3+1can be written as the sum of an even function g(x) and an odd function h(x).Then the value of g(0) is
g(x)=f(x)+f(-x)2 =12 x+1x3+1+1-xx3+1 =121x2-x+1+11+x+x2 =122(x2+1)(x2+1)2-x2 ∴g(0) = =1