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Critical Points and Extreme Values of f(x)

Find the critical points and absolute maximum/minimum of the given function f(x) in the specified intervals.

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Critical Points and Extreme Values of f(x)

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  1. Find the critical point of f(x) =(2x - 1)2 - 3 on [0, 4] • x = ½ • x = 1 • x = 2 • x = 3

  2. Find the critical point of f(x) =(2x - 1)2 - 3 on [0, 4] • x = ½ • x = 1 • x = 2 • x = 3

  3. Where is the absolute min of f(x) =(2x - 1)2 - 3 on [0,3] • x = 0 • x = ½ • x = 1 • x = 2 • x = 3

  4. Where is the absolute min of f(x) =(2x - 1)2 - 3 on [0,3] • x = 0 • x = ½ • x = 1 • x = 2 • x = 3

  5. Find the critical point of f(x) =sin2(x)-x on [-p/2, p/2]

  6. Find the critical point of f(x) =sin2(x)-x on [-p/2, p/2] f ’(x) = 2sin(x)

  7. Find the critical point of f(x) =sin2(x)-x on [-p/2, p/2] f ’(x) = 2sin(x)cos(x)

  8. Find the critical point of f(x) =sin2(x)-x on [-p/2, p/2] f ’(x) = 2sin(x)cos(x)-1=0

  9. Find the critical point of f(x) =sin2(x)-x on [-p/2, p/2] 2sin(x)cos(x)-1=0 2tan(x) = sec2(x)

  10. Find the critical point of f(x) =sin2(x)-x on [-p/2, p/2] 2sin(x)cos(x)-1=0 2tan(x) = sec2(x) 2tan(x) = tan2 (x) + 1 sin2 x + cos2 x = 1 = sec2(x) cos2x cos2 x cos2x

  11. Find the critical point of f(x) =sin2(x)-x on [-p/2, p/2] 2sin(x)cos(x)-1=0 2tan(x) = sec2(x) 2tan(x) = tan2 (x) + 1 0 = tan2 (x)-2tan(x) + 1 sin2 x + cos2 x = 1 cos2x cos2 x cos2x

  12. Find the critical point of f(x) =sin2(x)-x on [-p/2, p/2] 2sin(x)cos(x)-1=0 2tan(x) = sec2(x) tan2 (x)-2tan(x)+1=0 (tan(x) – 1)2 = 0

  13. Find the critical point of f(x) =sin2(x)-x on [-p/2, p/2] 2sin(x)cos(x)-1=0 2tan(x) = sec2(x) tan2 (x)-2tan(x)+1=0 (tan(x) – 1)2 = 0 tan(x) = 1

  14. tan(x) = 1f(x) =sin2(x)-x on [-p/2, p/2] • x = p/6 • x = p/4 • x = p/3

  15. Find the critical point of f(x) =sin2(x)-x on [-p/2, p/2] • x = p/6 • x = p/4 • x = p/3

  16. The only critical point of f(x)=sin2(x) - x on [-p /2, p /2] is x= p/4.Where is the absolute max? • x = p/2 y = 1 - p/2 • x = p/4 y = ½ - p/4 • x = -p/2 y = 1 + p/2

  17. The only critical point of f(x)=sin2(x) - x on [-p /2, p /2] is x= p/4.Where is the absolute max? • x = p/2 y = 1 - p/2 • x = p/4 y = ½ - p/4 • x = -p/2 y = 1 + p/2

  18. Extreme Value Theorem If f is continuous on [a, b], then f has an absolute maximum and an absolute minimum on [a, b].

  19. Extreme Value Theorem If f has an extreme value on an open interval, it must occur at a critical point. Remember c is a critical point iff f’(c) d.n.e. or f’(c) = 0.

  20. Rolle’s Theorem If f : [a, b] -> K is continuous on [a, b], differentiable on (a, b), and f(a) = f(b) = 0, then there exists a c (a, b) such that f '(c) = 0.

  21. proof: Case 1: f(x) = 0 on [a, b]. In this case, f '[(a+b)/2] = 0 and (a+b)/2 is inbetween a and b.

  22. proof: case 2: f(x) > 0 for some x in (a, b). Since f is continuous on [a, b], the extreme value theorem states that there is a c in [a, b] such that f has a maximum at c. Because f is 0 at a and b, this maximum must occur between a and b. Since all maximums are critical points and f is differentiable on (a, b), f '(c) = 0.

  23. proof:case 3: f(x) < 0 for some x in (a, b).

  24. Show f(x) =(2x+3)2 - 9 on [-3,0] satisfies the hypothesis of Rolle’s theorem. f is a polynomial so • f is continuous on [-3,0], and f is differentiable on (-3,0). • f(-3) = 9 – 9 = 0 • f(0) = 9 - 9 = 0

  25. f(x) =(2x+3)2 – 9 f’(x)= • f’(x) = 2(2x+3)2 • f’(x) = 2(2x+3) • f’(x) = 2(2)

  26. f(x) =(2x+3)2 – 9 f’(x)= • f’(x) = 2(2x+3)2 • f’(x) = 2(2x+3) • f’(x) = 2(2)

  27. f’(x) = 2(2x+3)2 f has a critical pt when x = • 2/3 • -2/3 • -3/2

  28. f’(x) = 2(2x+3)2 f has a critical pt when x = • 2/3 • -2/3 • -3/2

  29. Find the c in [-3,0] that satisfies the conclusion of Rolle’s theorem. f(x) =(2x+3)2 - 9 f’(x) = 2(2x+3) • 4(2x+3) = 0 when • x = - 3/2 therefore f’(-3/2) = 0 and -3/2 (-3, 0) which means -3<-3/2<0

  30. Prove y = x3 – 8x - 3 has a zero in [-1, 1]. y(-1) = 4 > 0 y(1) = -10 < 0 I.V.T. guaranties a d (-1, 1) so that f(d) = 0 because f is continuous on [-1, 1]

  31. y = x3 – 8x – 3dy/dx = • 3x - 8 • 3x2 - 8 • 3x2 - 8x - 3

  32. y = x3 – 8x – 3dy/dx = • 3x - 8 • 3x2 - 8 • 3x2 - 8x - 3

  33. Prove y = x3 – 8x - 3 has exactly one zero in [-1, 1]. Proof by contradiction Suppose y(a) = y(b) = 0 and {a,b} (-1, 1) and a < b.

  34. Prove y = x3 – 8x - 3 has exactly one zero in [-1, 1]. If 2, by Rolle’s theorem, there exists a c (a, b) (-1,1) such that y’(c) = 0. y’ = 3x2 - 8 = 0 when x =

  35. Mean Value Theorem If f : D -> K is continuous on [a, b], differentiable on (a, b), then there exists a c (a, b) such that

  36. proof: Let g : [a, b] -> R be the straight line function passing through (a, f(a)) and (b, f(b)). This means that g(x) =

  37. proof: g(x) = m(x - a) + f(a) where

  38. proof: Next, define a new function k : [a, b] -> R by k(x) = f(x) - g(x). Note that : k is continuous on [a, b], k is differentiable on (a, b). I want to apply Rolle’s Theorem

  39. proof: k(x) = f(x) - g(x). k(x) = f(x) - m(x - a) - f(a) k(a) = f(a) - m(0) – f(a)= 0. k(b) = f(b) –m(b –a) – f(a) = f(b) – [f(b) – f(a)] – f(a) = 0

  40. proof: Thus by Rolle’s Theorem, there exists a c in (a, b) such that k'(c) = 0. k(x) = f(x) - m (x-a) - f(a) Differentiating gives k'(x) =

  41. proof: k(x) = f(x) - m (x-a) - f(a) Differentiating gives k'(x) = f '(x) – m and replacing x by c gives

  42. Mean Value Theorem If f : D -> K is continuous on [a, b], differentiable on (a, b), then there exists a c in (a, b) such that .

  43. Can we apply the mean value theorem for on [1, 5]? Continuous on [1, 5]? Differentiable on (1, 5)? Then there exists a c in (1, 5) such that .

  44. Find f’(x) Note 2.5 is in (1, 5) so c = 2.5.

  45. Let f(x) = x3 + 3x + 1 on [1, 2]. Find f ’(x). • 3x + 3 • 3x3 + 3x • 3x2 + 3

  46. Let f(x) = x3 + 3x + 1 on [1, 2]. Find f ’(x). • 3x + 3 • 3x3 + 3x • 3x2 + 3

  47. f(x) = x3 + 3x + 1 find • 8 • 10 • 15 • 20

  48. f(x) = x3 + 3x + 1 find • 8 • 10 • 15 • 20

  49. Where does 3x2 + 3 = 10on [1, 2]?

  50. Where does 3x2 + 3 = 10on [1, 2]? • 1.53 • 0.1

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