Solve the equation by graphing: x^2+5x-14

Which value is not a solution of 8x^3 – 1 = 0

Which of the following is the result of expanding

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.