Explanation Given information: The function is f ( x ) = x m ( 1 − x ) n , where m , n ≥ 2 are integers. Definition used: For any function f , a point p in the domain of f where f ' ( p ) = 0 or f ' ( p ) is undefined is called a critical point of the function. Concept used: Second derivative test: If f ' ' is positive, f has a local minimum at the critical point. If f ' ' is negative, f has a local maximum at the critical point. Calculation: Let us consider the given function f ( x ) = x m ( 1 − x ) n . Differentiating with respect to ‘ x’, we have f ' ( x ) = m x m − 1 ( 1 − x ) n + n x m ( 1 − x ) n − 1 ( − 1 ) = m x m − 1 ( 1 − x ) n − n x m ( 1 − x ) n − 1 Equate f ' ( x ) = 0 to find critical points. ⇒ f ' ( x ) = 0 ⇒ m x m − 1 ( 1 − x ) n − n x m ( 1 − x ) n − 1 = 0 ⇒ x m − 1 ( 1 − x ) n − 1 ( m ( 1 − x ) − n x ) = 0 ⇒ x m − 1 ( 1 − x ) n − 1 ( m − m x − n x ) = 0 ⇒ x m − 1 ( 1 − x ) n − 1 ( m − x ( m + n ) ) = 0 ⇒ x m − 1 = 0 o r ( 1 − x ) n − 1 = 0 o r ( m − x ( m + n ) ) = 0 ⇒ x = 0 o r x = 1 o r x = m m + n The critical points are x = 0 , x = 1 and x = m m + n . Hence, the critical points are x = 0 , x = 1 and x = m m + n

Applied Calculus, 6e WileyPLUS + Loose-leaf

6th Edition

ISBN: 9781119408901

Author: Deborah Hughes-Hallett, Andrew M. Gleason, Patti Frazer Lock, Daniel E. Flath

Publisher: Wiley (WileyPLUS Products)

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Chapter 4.1, Problem 58P

To determine

To find:

The classification of all critical points of. f(x)=xm(1−x)n.

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Chapter 4.1, Problem 57P

Chapter 4.2, Problem 1P

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Applied Calculus, 6e WileyPLUS + Loose-leaf

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