, x ∊ [- 1 , 1] We have to find the maximum value of f (x) Now, Differentiating f (x) w.r.t x, we get f'(x) =
(4x+x2)−(2x+1)x
(4+x+x2)2
=
4+x+x2−2x2−x
(4+x+x2)2
=
4−x2
(4+x+x2)2
⇒ f'(x) > 0 for all x ∊ (- 2 , 2) ⇒ In the interval (-2, 2), f (x) is increasing ⇒ f (x) is also increasing in [-1 , 1] ⊂ (-2 , 2) ⇒ f (x) is maximum at x = 1 ∴ fmax = f (1) =