Let the function f (x) = 3 sin x – 4 cos x + log (|x| + 1+x2 be defined on the interval [0, 1]. The odd extension of f (x) to the interval [– 1, 1] is

To make f (x) an odd function in the interval [–1, 1], we re-define f (x) as follows :f(x)=f(x),    0≤x≤1−f(−x),    −1≤x<0=3sin⁡x−4cos⁡x+log⁡|x|+1+x2,    0≤x≤1−−3sin⁡x−4cos⁡x+log⁡|x|+1+x2,    −1≤x<0=3sin⁡x−4cos⁡x+log⁡|x|+1+x2,0≤x≤13sin⁡x+4cos⁡x−log⁡|x|+1+x2,−1≤x<0Thus, the odd extension of f (x) to the interval [–1, 1] is3sin⁡x+4cos⁡x−log⁡|x|+1+x2