The curve 4 y=3 x^4-2 x^2 attains -------- at the points x=- 1 3 and x= 1 3

Step-by-step solution

Given the curve 4y = 3x^4 - 2x^2 , we have y = 3 4 x^4 - 1 2 x^2 . To find the critical points, we calculate the first derivative with respect to x : dy dx = 3 4 (4x^3) - 1 2 (2x) = 3x^3 - x = x(3x^2 - 1) . Setting dy dx = 0 , we get x = 0 , x = 1 3 , and x = - 1 3 . To determine the nature of these points, we calculate the second derivative: d^2y dx^2 = 9x^2 - 1 . Evaluating at x = 1 3 : d^2y dx^2 = 9 ( 1 3 ) - 1 = 3 - 1 = 2 . Since d^2y dx^2 > 0 at both x = 1 3 and x = - 1 3 , the function attains a local minimum