Solve 2x2 + 12x − 14 = 0 by completing the square, options: a. 2(x + 6)2 = 26; x = −6 + square root 13, x = −6 − square root 13 b. 2(x + 6)2 = 20; x = −6 + square root 20, x = −6 − square root 20 c. 2(x + 3)2 = 32; x = 1, x = −7 d. 2(x + 3)2 = 32; x = 13, x = 19

Does the function satisfy the hypotheses of the mean value theorem on the given interval? f(x) = 4x^2 + 3x + 4, [−1, 1] no, f is continuous on [−1, 1] but not differentiable on (−1, 1). no, f is not continuous on [−1, 1]. yes, f is continuous on [−1, 1] and differentiable on (−1, 1) since polynomials are continuous and differentiable on . there is not enough information to verify if this function satisfies the mean value theorem. yes, it does not matter if f is continuous or differentiable; every function satisfies the mean value theorem.