CONCAVE UPWARDS

1. CONCAVE UPWARDS g"(x) > 0

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

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

4. EXAMPLE 1 Determine the maximum, minimum and inflection points of the function f(x) = x3 – 3x2 – 9x + 15 Step 1: Find the first and second derivatives f ′ (x) = 3x2 – 6x – 9 f " (x) = 6x – 6 Step 2: Find critical values of x by setting the first derivative equal to 0. 3x2 – 6x – 9 = 0 x2 – 2x – 3 = 0 (x + 1) (x – 3) = 0 x = –1 or x = 3

5. Step 3: Use the second derivative to determine maximum or minimum f " (x) = 6x – 6 f " (–1) = 6(–1) – 6= –12 f(–1) = (–1)3 – 3(–1)2 – 9(–1) + 15 = 20 P(–1, 20) is a maximum pointon the graph 20 is the maximum value f " (3) = 6(3) – 6 = 12 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. EXAMPLE 1 continued Step 4: Set the second derivative equal to zero and solve for the potential inflection point:. f " (x) = 6x – 6 f " (0) = - 6 f " (2) = 6 6x – 6 = 0 CD CU x = 1 1 Since the concavity changes at x = 1, we know that there is an inflection point at this value Step 5: Find the yvalue of the point. f(1) = (1)3 – 3(1)2 – 9(1) + 15 = 4 IP (1, 4)

7. (1, 4)

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. 0 2 Step 3: Test where the second derivative is positive or negative. g"(-1) = 9 g"(1) = -3 g"(3) = 9 CU CD CU 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. (0, 5) (2, -3)

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

12. p/2 Step 3: Test where the second derivative is positive or negative. h“(x) = cosx CAST Rule All Sin Tan Cos + + 0 CU 2p CU CD 3p/2 Since the concavity changes at bothp/2and 3p/2 there are inflection points at these x-values. h(p/2) = p/2and h(3p/2) = 3p/2 The inflection points are at (p/2, p/2)and (3p/2, 3p/2)