Extremum and inflection of point

1. Extremum and inflection of point Presented by:- :: Bothaina AL-sobai :: :: Meji Le :: :: Hind Nader :: :: Hanan :: ::Aisha alsulati ::

2. Outline… • Definition (Hanan) • Extreme Value Theorem (Hanan) • Critical Number (Hanan) • The First Derivative Test (Bothaina) • The Second Derivative Test (Hind) • Inflection Point (MiJi)

3. Definition of Extremum

4. Extreme Value Theorem States that if a function f(x) is continuous in the closed interval [a, b] then f(x) must attain its maximum and minimum value, each at least once.

5. Extreme Value Theorem Maximum Minimum

6. Critical Number Critical number means: If a number c is in the domain of a function f, then c is called a critical number of provided f ' (c) = 0 or f ‘ (c) doesn't exist.

7. Example (1) • Find the critical number for the function f(x) = • F '(x) = • Then f '(x) is zero at x=3, and x=-1 and f ' is undefined at x=1, however x=1 is not in the domain of f and hence the only critical numbers of f are x=3 and x=-1 • SO the critical point (3,6) and(-1,-2)

8. Keep in mind….

9. The First Derivative Test To find extrema we should follow 3 steps: 1)Find the derivative of f and set the equation to zero. 2) Find the critical points 3)Check the signs (by taking test points )

10. Example (2) • Locate all relative extrema for the function

11. Example (2) • Step number 1,2 F '(x)= Step number 1 = = Step number 2 The only critical numbers of f are -2 and 3.

12. Third step… • Using theorem we conclude that the critical number -2 yields a relative maximum (f ' changes sign from + to -) and 3 yields a relative minimum

13. Remember… • f '(X) changes sign from + to -). • f '(x) changes sign from – to +). MAX(-2) MIN(3)

15. The Second Derivative Test Suppose f'(x)= 0 • If f"(x) < 0, then f has a relative maximum at x • If f"(x) > 0, then f has a relative minimum at x

16. Example (3) 1) f(x)= 18x – (2/3)x³ f'(x)= 18 – 2x² = 0 2 (9 – x2) = 0 2 (3 – x) (3 + x) = 0 x = 3, -3 f'' (x)= -4x when x = 3, y'' = -4(3) = -12 < 0 so there is a relative maximum when x = -3, y'' = -4(-3) = 12 > 0 so there is a relative minimum

17. Example (4) • f(x)= 6x4 – 8x³ + 1 f' (x) = 24x³ – 24x² = 0 24x² (x – 1) = 0 x = 0, 1 f" (x) = 72x² – 48x

18. when x = 1, f'' (x)= 72(1) – 48(1) = 24 > 0 so there is a relative minimum when x = 0we can't apply the second derivative test when x = 0 if x < 0 then f' (x)< 0 if 0 < x < 1, then f' (x)< 0thus, no maximum or minimum exist when x=0

19. Inflection Point

20. D2f f Tangent at a point of inflexion. P A point on a curve at which the tangent crosses the curve itself.

21. Inflection Point • A point on a curve at which the slope of graph does not changes

22. Property of point of inflection • An important point on a graph is one which marks a transition between a region where the graph is concave up and one where it is concave down. We call such a point an inflection point. • If (c,f(c)) is a point of a inflection of f, then either f’’(c) =0 or does not exist.

23. Example(6) F(x) = 6x4 – 8x3 + 1 • Step 1. f' (x) = 24x³ – 24x² • Step 2. f'' (x) = 72x² – 48x = 24x (3x – 2) • Step 3. To find where f''(x)= 0 24x (3x - 2) = 0 24x = 0 OR 3x – 2 = 0 X = 0 x = 2/3

25. Step 4. Check on ConcavitySign of second derivative f'' (x)= 24x (3x – 2) f’’ (-1)= 24x (3x – 2) = 24(-1) (3(-1) – 2) = -24 (-3 -2) = -24 (-5) = 120 f’’ (0.1)= 24(0.1) (3(0.1) – 2) = 2.4 (0.3 – 2) = 2.4 (-1.7) = -4.08

26. Step 5. Check on slope f' (x) = 24x³ – 24x² f‘(-1) = 24(-1)³ – 24(-1)² = -24 – 24 = -48 f‘(0.1) = 24x³ – 24x² = 24(0.1)³ – 24(0.1)² = 0.024 – 0.24 = -0.216

27. THANK YOU! For Your Attention