local max. value is 0 at each of the points x = 1 and x = −1 and local min. value is -3456/3125 at x = -1/5
F’(x) = -(x -1)32(x+1) - 3(x -1)2(x+1)2 = 0
⇒ x = 1, -1, -1/5
Since, f || (1) and f || (-1)<0, 1 and -1 are the point of local max.
F || (-1/5)>0, -1/5 is the point of local min.
F(1) = f(-1) = 0
Also, f(-1/5) = -3456/3125