The denominator of a fraction exceeds the square of the numerator by 16, then the least value of the fraction is

Step-by-step solution

Let the numerator be x , then f x = x x 2 + 1 6 On differentiating w.r.t. x , we get f ′ x = x 2 + 1 6 · 1 - x 2 x x 2 + 1 6 2 = x 2 + 1 6 - 2 x 2 x 2 + 1 6 2 = 1 6 - x 2 x 2 + 1 6 2 ...(i) Putting f ' ( x ) = 0 for maxima or minima, we get, 16 - x 2 = 0   ⇒   x = 4 , - 4 Again, on differentiating w.r.t. x , we get f ″ x = x 2 + 1 6 2 - - 2 x - 1 6 - x 2 2 x 2 + 1 6 2 x x 2 + 1 6 4 At x = 4 ,   f " ( x ) < 0 ∴ f ( x ) is maximum at x = 4 And at x = - 4 ,   f