1. 1 CONCAVE UPWARDS g"(x) > 0

2. 2 CONCAVE DOWNWARDS g"(x) < 0 negative slope y = g(x) positive slope zero slope

3. 3 POINTS OF INFLECTION If a change in concavity occurs, then an inflection point exists. CU CD

4. 4 EXAMPLE 1 Determine the maximum, minimum and inflection points of the function f(x) = x 3 – 3x 2 – 9x + 15 Step 1: Find the first and second derivatives f ′ (x) = 3x 2 – 6x – 9 f " (x) = 6x – 6 Step 2: Find critical values of x by setting the first derivative equal to 0. 3x 2 – 6x – 9 = 0 x 2 – 2x – 3 = 0 (x + 1) (x – 3) = 0 x = –1 or x = 3

5. 5 Step 3: Use the second derivative to determine maximum or minimum f " (x) = 6x – 6 f " (–1) = 6(–1) – 6 = –12 f " (3) = 6(3) – 6 = 12 f(–1) = (–1) 3 – 3(–1) 2 – 9(–1) + 15 = 20 P(–1, 20) is a maximum point on the graph 20 is the maximum value f(3) = (3) 3 – 3(3) 2 – 9(3) + 15 = –12 P(3, –12) is a minimum point on the graph -12 is the minimum value

6. 6 EXAMPLE 1 continued Step 4: Set the second derivative equal to zero and solve for the potential inflection point:. f " (x) = 6x – 6 6x – 6 = 0 x = 1 1 f " (0) = - 6 f " (2) = 6 CDCU Since the concavity changes at x = 1, we know that there is an inflection point at this value Step 5: Find the y value of the point. f(1) = (1) 3 – 3(1) 2 – 9(1) + 15 = 4 IP (1, 4)

7. 7 (1, 4)

8. 8 EXAMPLE 2 Find the concavity intervals and any points of inflection for the function Step 1: Find the first and second derivatives Step 2: Set the second derivative equal to zero and solve for the potential inflection points x = 0 or x = 2

9. 9 Step 3: Test where the second derivative is positive or negative. 02 g"(-1) = 9g"(1) = -3g"(3) = 9 CUCDCU Since the concavity changes at both 0 and 2 there are inflection points at these x -values. g(0) = 5 and g(2) = -3 The infection points are at (0, 5) and (2, -3)

10. 10 (0, 5) (2, -3)

11. 11 EXAMPLE 3 Find the concavity intervals and any points of inflection for the function h(x) = x – cos x on the interval (0, 2  ) Step 1: Find the first and second derivatives Step 2: Set the second derivative equal to zero and solve for the potential inflection points h’(x) = 1 + sin x h“(x) = cos x cos x = 0 x =  / 2 or x = 3  / 2

12. 12 Step 3: Test where the second derivative is positive or negative. h“(x) = cos x /2/2 CUCD CU  / 2 CAST Rule All Sin TanCos 0 22 + + Since the concavity changes at both  / 2 and  / 2 there are inflection points at these x -values. h(  / 2 ) =  / 2 and h(  / 2 ) =  / 2 The inflection points are at (  / 2,  / 2 ) and (  / 2,  / 2 )

13. 13