C.5 Concavity and the Second Derivative Test

1. C.5 Concavity and the Second Derivative Test

2. Concavity The graph of a differentiable function y = f(x) is • Concave up on an open interval I if y’ is increasing on I • Concave down on an open interval I if y’ is decreasing on I • If a function y = f(x) has a second derivative, then we can conclude that y’ increases if y” > 0 and y’ decreases if y” < 0.

3. Concavity Test • The graph of a twice differentiable function y = f(x) is: • Concave up on any interval where y” > 0 • Concave down on any interval where y” < 0

4. Example 1 Determining Concavity Use the Concavity Test to determine the concavity of the given function on the given interval. Since the second derivative is always negative regardless of the interval, the function is concave down on any interval.

5. Point of Inflection A point where the graph of a function has a tangent line and where the concavity changes is a point of inflection. Given a function f that is continuous, If f ”(c) = 0, then (c, f(c)) is a candidate for an inflection point. • If f ” (x) changes sign as x increases through c, then (c, f(c)) is an inflection point. • If f ” (x) does not change sign as x increases through c, then (c, f(c)) is not an inflection point.

6. Example 2 Determining Concavity + - y = 3 + sinx on (0, 2π) y’ = cos x y “ = -sin x 2π 0 π Concave up Concave down Point of Inflection

7. Drill: use the concavity test to determine where the graph is concave up or down. Drill: use the first derivative test to determine intervals of increasing and decreasing and the local extrema. • y = x2 – x – 1 • y’ = 2x -1 • 2x – 1 = 0 • X = ½ • Increasing ( ½ , ∞) • Decreasing (-∞, ½ ) • @ x = ½, there is a local minimum, according to test • (1/2, -5/4) is the minimum. • y = 4x3 + 21x2 + 36x -20 • y ‘ = 12x2 + 42x + 36 • y” = 24x + 42 • 24x + 42 = 0 • x = -7/4 • Concave up: (-7/4, ∞) • Concave down: (-∞. -7/4) - + 1/2 - + -7/4

8. Example: Studying Motion Along a Line • A particle is moving along the x-axis with the position function s(t) = 2t3 – 14t2 + 22t – 5, t> 0. Find the velocity and acceleration, and describe the motion of the particle.

9. A particle is moving along the x-axis with the position function s(t) = 2t3 – 14t2 + 22t – 5, t> 0. Find the velocity and acceleration, and describe the motion of the particle. • Velocity: 6t2 – 28t + 22 2(3t2 – 14t + 11) = 0 2(3t-11)(t-1) = 0 Intervals: + right increasing - left decreasing + increasing right

10. Acceleration • A(t) = 12t – 28 • 0 = 12t – 28 • t = 7/3 - Decel-erating Concave down + Concave up Accel- erating

11. Homework C.5 • Pg. 195 • #1, 2, 11, 14, 16 • #27 On this one find, intervals of increasing and decreasing, relative extrema, concavity, and points of inflection. • Extra problem A particle is moving along the x-axis with position function x(t) = 3t² - 2t³ Find the Velocity and Acceleration and describe the motion of the particle for t ≥ 0.