AP Calculus AB Final: Multiple Choice Questions And Answers

AP Calculus AB Final: Multiple Choice Questions Find all points of relative minima and maxima: y = x^2 + 6x + 5 - Answer-minima at (-3, -4) Find all points of relative minima and maxima: y = -12/(x^2 + 3) - Answer-minima at (0, -4) Find the open intervals where the function is concave up and concave down. y = (2x-12)^(1/3) - Answer-concave up: (-∞, 6) concave down: (6, ∞) Find the open intervals where the function is concave up and concave down. y = -x^3 + x^2 + 3 - Answer-concave up: (-∞, 1/3) concave down: (1/3, ∞) Find x-coordinates of all points of inflection. y = -4/(x^2 + 4) - Answer-POIs: x = -2/√(3), 2/√(3) Find open intervals where function is increasing and decreasing. f(x) = (6x^2 - 6)/x^3 - Answer-inc: (-√(3), 0)∪(0, √(3)) dec: (-∞, -√(3))∪(√(3), ∞) Find open intervals where function is increasing and decreasing. f(x) = -[(x+2)/(x+3)]^3 - Answer-inc: (-3, -2) dec: (-∞, -3)∪(-2, ∞) Find the x-coordinates of all critical points. f(x) = -x^3 + x^2 + 4 - Answer-x = 0, 2/3 Evaluate each limit using L'Hopital's Rule. limx-0(2x + 1)^(1/x) - Answer-e^2 Evaluate each limit using L'Hopital's Rule. limx-π(4secx - 4tanx) - Answer-DNE Find the values of c that satisfy the Mean Value Theorem. f(x) = 2x^2 - 16x + 30; [4, 6] - Answer-c = 5 Find the values of c that satisfy the Mean Value Theorem. f(x) = x^2/(4x + 8); [-1, 3] - Answer-c = -2 + √(5), -2 - √(5) A particle moves along a horizontal line. Its position function is s(t) for t = 0. Find the velocity function v(t). s(t) = t^3 - 14t^2 - Answer-v(t) = 3t^2 - 28t A particle moves along a horizontal line. Its position function is s(t) for t = 0. Find the acceleration function a(t). s(t) = -t^3 + 9t^2 - Answer-a(t) = -6t + 18 A particle moves along a horizontal line. Its position function is s(t) for t = 0. Find the times t when the particle changes directions and times t when the acceleration is 0. s(t) = -t^3 + 26t^2 - 169t - Answer-a = 0 at t = 26/3 particle changes direction at t = 13/3, 13 Find equation of the line tangent to the function at the given point. y = -ln(x + 1) at (2, -ln3) - Answer-y = -(1/3)x + 2/3 - ln3 Find equation of the line tangent to the function at the given point. y = 2cos(x) at (π/6, √(3)) - Answer-y = -x + π/6 + √(3) Find equation of the line tangent to the function at the given point. y = -3/(x^2 + 1) at (3, -3/10) - Answer-y = (9/50)x - 42/50 Find the indicated derivative with respect to x. y = 5x^(5/4) + 2x^(1/3) + 4x^(1/5); Find (d^2*y)/(d*x^2) - Answer-(d^2*y)/(d*x^2) = 25/(16x^(3/4)) - 4/(9x^(5/3)) - 16/(25*x^(9/5)) Find the indicated derivative with respect to x. y = 2/x^2; Find (d^2*y)/(d*x^2) - Answer-(d^2*y)/(d*x^2) = 12/x^4 Differentiate with respect to x. y = cos^-1(2x^2) - Answer-dy/dx = -4x/√(1-4x^4) Differentiate with respect to x.