. . . (i) On differentiating w.r.t. x, we get ‌f′(x)=‌
(x2+x+1)(2x−1)−(x2−x+1)(2x+1)
(x2+x+1)2
‌‌ for maximum or minimum, put ‌f′(x)=0 ‌⇒(x2+x+1)(2x−1)−(x2−x+1)(2x+1)=0 ‌⇒‌‌x2+x−1−(−x2+x+1)=0 ‌⇒‌‌2x2−2=0⇒x=±1 ‌‌ Now, ‌f′(c)=‌
2x2−2
(x2+x+1)2
Again differentiating, we get ‌(x2+x+1)2(4x)−(2x2−2) ‌f′′(x)=‌
2(x2+x+1)(2x+1)
(x2+x+1)4
at x=1,f′′(x)>0 Therefore it is minimum at x=1 Put x=1 in equation (i), we get f(1)=‌