Let the function `f(x)=x^2+x+s in x-cosx+log(1+|x|)` be defined on the interval `[0,1]` .Define functions `g(x)a n dh(x)in[-1,0]` satisfying `g(-x)=-f(x)a n dh(-x)=f(x)AAx in [0,1]dot`

Answer

`g(x)= -x^(2)+x+sinx +cosx -log(1+|x|)`
`h(x)=x^(2)-x-sinx-cosx+log(1+|x|)`

Explanation

Clearly `g(x)` is the odd extension of the function `f(x)` and `h(x)` is the even extension.
Since `x^(2),cosx, log(1+|x|)` are even functions and `x, sin x` and odd functions.
`g(x)= -x^(2)+x+sinx +cosx-log(1+|x|)`
and `h(x)=x^(2)-x-sinx-cosx +log(1+|x|)`
Clearly this function satisfies the restriction of the problem.