If y = _ e x^3+3 ⁻¹ x+ k x^2 and y ^ ( 1 2 )=2 3 , then k =

Step-by-step solution

Given y = _e x^3 + 3 ⁻¹ x + k x^2 , simplifying the logarithm yields y = 3 _e x + 3 ⁻¹ x + k x^2 . Differentiating, y' = 3 1 x + 3 1 1 - x^2 + 2k x = 3 x + 3 1 - x^2 + 2k x . At x = 1 2 , y' ( 1 2 ) = 3 1/2 + 3 1 - ( 1 2 )^2 + 2k 1 2 = 6 + 3 3/4 + k = 6 + 3 3 /2 + k = 6 + 6 3 + k . Rationalizing, 6 3 = 2 3 , so y' ( 1 2 ) = 6 + 2 3 + k . Given y' ( 1 2 ) = 2 3 , equating gives 6 + 2 3 + k = 2 3 , hence k = -6 . B