Explanation Given: f ( x , y ) = x 3 + 2 x y − 2 y 2 − 10 x Let P = ( a , b ) be a critical point of f ( x , y ) . Let the second order derivatives f x x , f y y , f x y be continuous near P . Let D be the discriminant of the function f ( x , y ) and is defined as D ( x , y ) = f x x f y y − f x y 2 (a) If D > 0 , then f ( a , b ) is a local minimum or local maximum. (b) If D < 0 , then ( a , b ) is a saddle point. (c) If D = 0 , then the test is inconclusive. (d) If D > 0 and f x x ( a , b ) > 0 then f ( a , b ) is a local minimum. (a) If D > 0 and f x x ( a , b ) < 0 then f ( a , b ) is a local maximum. Calculation: f ( x , y ) = x 3 + 2 x y − 2 y 2 − 10 x By differentiating, we get, f x = 3 x 2 + 2 y − 10 f y = 2 x − 4 y For critical points putting f x = 0 , f y = 0 3 x 2 + 2 y − 10 = 0 and 2 x − 4 y = 0 3 ( 2 y ) 2 + 2 y − 10 = 0 ( Putting the value of x ) 12 y 2 + 2 y − 10 = 0 6 y 2 + y − 5 = 0 6 y 2 + 6 y − 5 y − 5 = 0 6 y ( y + 1 ) − 5 ( y + 1 ) = 0 y = − 1 , 5 6 x = 2...

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Author: Rogawski

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Chapter 14.7, Problem 9E

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The critical points of the function and use the Second Derivative Test to determine whether they are local minimum, local maximum, saddle points or state that the test fails.

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Chapter 14.7, Problem 8E

Chapter 14.7, Problem 10E

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Chapter 14 Solutions

CALCULUS:EARLY TRANS.-SAPLINGPLUS

Ch. 14.1 - Prob. 1PQ Ch. 14.1 - Prob. 2PQ Ch. 14.1 - Prob. 3PQ Ch. 14.1 - Prob. 4PQ Ch. 14.1 - Prob. 5PQ Ch. 14.1 - Prob. 1E Ch. 14.1 - Prob. 2E Ch. 14.1 - Prob. 3E Ch. 14.1 - Prob. 4E Ch. 14.1 - Prob. 5E

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The critical points of the function and use the Second Derivative Test to determine whether they are local minimum , local maximum , saddle points or state that the test fails.

The critical points of the function and use the Second Derivative Test to determine whether they are local minimum, local maximum, saddle points or state that the test fails.

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Chapter 14 Solutions