Option : (C)
f(x) = \(\frac{x}{4+x+x^2}\), x ∈ [-1,1]
Differentiating f(x) with respect to x, we get
Since,
4 - x2 > 0 ∀ x∈[-1,1] and
(4 + x + x2)2 > 0 ∀ x ∈ R
Therefore,
f’(x) > 0 ∀ x ∈ [-1,1]
Hence,
f(x) is increasing in [-1,1] and therefore the maximum value of f(x) occurs at x = 1 and
f(1) = \(\frac{1}{6}\)