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Post a LessonAnswered on 27/01/2022 Learn Unit III: Calculus/Applications of Derivatives
Pugazhendhi V
IT Technical Executive with 4 years of experience
Solution:
Answered on 27/01/2022 Learn Unit III: Calculus/Applications of Derivatives
Pugazhendhi V
IT Technical Executive with 4 years of experience
Solution:
Answered on 27/01/2022 Learn Unit III: Calculus/Applications of Derivatives
Pugazhendhi V
IT Technical Executive with 4 years of experience
Solution:
Answered on 27/01/2022 Learn Unit III: Calculus/Applications of Derivatives
Pugazhendhi V
IT Technical Executive with 4 years of experience
Solution:
Given; f(x) = (x + 1)3 (x – 3)3 ⇒ f'(x) = (x + 1)3 3(x – 3)2 + (x – 3)33(x + 1)2 = 3(x + 1)2(x – 3)2(x + 1 + x – 3) = 3(x + 1)2(x – 3)2(2x – 2) = 6(x +1)2 (x – 3)2 (x -1) ⇒ 6(x +1)2 (x – 3)2 (x – 1) = 0 ⇒ x = -1, 1, 3 The intervals are (-∞, -1), (-1, 1), (1, 3), (3, ∞) f'(-2) = (-2 – 1) < 0 ∴ Strictly decreasing in (-∞, -1) f'(0) = (0 – 1) < 0 ∴ Strictly decreasing in (-1, 1) f'(2) = (2 – 1) > 0 ∴ Strictly increasing in (1, 3) f'(4) = (4 – 1) > 0 ∴ Strictly increasing in (3, ∞)
read lessAnswered on 27/01/2022 Learn Unit III: Calculus/Applications of Derivatives
Pugazhendhi V
IT Technical Executive with 4 years of experience
Solution:
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