Explanation To verify the provided statement, consider a function f ( x , y ) = ( x 2 − 1 ) 2 + ( y − x − 1 ) 2 Now, apply the second partials test on this function to find the relative extrema. The partial test is: Let f have continuous second partial derivatives on an open region containing a point ( a , b ) for which f x ( a , b ) = 0 and f y ( a , b ) = 0 . Then the second derivative test discriminant is defined as: d = f x x ( a , b ) f y y ( a , b ) − [ f x y ( a , b ) ] 2 If d > 0 and f x x ( a , b ) > 0 , then f has a relative minimum at ( a , b ). If d > 0 and f x x ( a , b ) < 0 , then f has a relative maximum at ( a , b ). If d < 0, then ( a , b , f ( a , b )) is a saddle point. The test is inconclusive if d = 0. Thus, first order partial derivatives of this function will be: f x = ∂ ∂ x [ ( x 2 − 1 ) 2 + ( y − x − 1 ) 2 ] = 2 ( x 2 − 1 ) × 2 x + 2 ( y − x − 1 ) × ( − 1 ) = 4 x ( x 2 − 1 ) − 2 ( y − x − 1 ) And, the derivative with respect to y : f y = ∂ ∂ y [ ( x 2 − 1 ) 2 + ( y − x − 1 ) 2 ] = 0 + 2 ( y − x − 1 ) = 2 ( y − x − 1 ) Now, to find the critical points, equate the first order partial derivatives to zero. Then, 4 x ( x 2 − 1 ) − 2 ( y − x − 1 ) = 0 …...…... (1) And, 2 ( y − x − 1 ) = 0 …...…... (2) Put equation (2) in equation (1): 4 x ( x 2 − 1 ) − 0 = 0 x = 0 , − 1 , 1 Thus, by putting obtained values of x in equation (2), the critical points are obtained as For x = 0 2 ( y − 0 − 1 ) = 0 y = 1 Thus, the point is A ( 0 , 1 ) For x = 1 2 ( y − 1 − 1 )

Calculus: Early Transcendental Functions (MindTap Course List)

6th Edition

ISBN: 9781285774770

Author: Ron Larson, Bruce H. Edwards

Publisher: Cengage Learning

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Chapter 13.8, Problem 57E

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Whether the statement “Between any two relative minima of f, there must be at least one relative maximum of f ” is true or false.

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Calculus: Early Transcendental Functions (MindTap Course List)

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Whether the statement “Between any two relative minima of f , there must be at least one relative maximum of f ” is true or false.

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Whether the statement “Between any two relative minima of f, there must be at least one relative maximum of f ” is true or false.

Chapter 13 Solutions