Let f: R R is differentiable function having f(3)=3, f^ (3)= 1 27 and g ( x )= array ccc ₃^ f (x) 3 t ^2 x-3 dt & , & if x 3 ~K & , & if x=3 array . is continuous at x=3 , then K =

Step-by-step solution

Continuity requires _ x 3 g(x) = g(3) , so K = _ x 3 ₃^ f(x) 3t^2 ,dt x-3 . The limit is indeterminate ( 0/0 ), so apply L'Hôpital's Rule. Differentiate numerator and denominator separately. Denominator derivative: d dx (x-3) = 1 . Numerator derivative (via Leibniz rule): d dx ₃^ f(x) 3t^2 ,dt = 3(f(x))^2 f'(x) . K = _ x 3 3(f(x))^2 f'(x) 1 = 3(f(3))^2 f'(3) . Given f(3)=3 and f'(3)= 1 27 , so K = 3 9 1 27 = 1 . 1