Explanation Given: The given function is, f ( x , y ) = x 2 − x y + y 2 + 2 x + 2 y + 6 . Approach: “For a function z = f ( x , y ) , let f x x , f y y , and f x y all exist in a circular region contained in the xy -plane with center ( a , b ) . Further, let f x ( a . b ) = 0 and f y ( a . b ) = 0 . Define the number D , known as the discriminant, by D = f x x ( a , b ) ⋅ f y y ( a , b ) − [ f x y ( a , b ) ] 2 Then, a. f ( a , b ) is a relative maximum if D > 0 and f x x ( a , b ) < 0 . b. f ( a , b ) is a relative minimum if D > 0 and f x x ( a , b ) > 0 . c. f ( a , b ) is a saddle point (neither a maximum nor a minimum) if D < 0 . d. if D = 0 , the test gives no information.” Calculation: Consider the given function, f ( x , y ) = x 2 − x y + y 2 + 2 x + 2 y + 6 . Find the values, f x ( x , y ) , and f y ( x , y ) : f x ( x , y ) = ∂ ∂ x ( x 2 − x y + y 2 + 2 x + 2 y + 6 ) = 2 x − y + 2 and, f y ( x , y ) = ∂ ∂ y ( x 2 − x y + y 2 + 2 x + 2 y + 6 ) = − x + 2 y + 2 A function has relative extrema at a point, if f x ( x , y ) = 0 2 x − y + 2 = 0 And if f y ( x , y ) = 0 ,

2nd Edition

ISBN: 9780321964038

Author: GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Publisher: Pearson Addison Wesley,

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Chapter 9.3, Problem 5E

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All the points for relative extrema and identify any saddle point for the given function.

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Chapter 9.3, Problem 4E

Chapter 9.3, Problem 7E

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The graph of a function f is given. The x y-coordinate plane is given. A curve with 3 parts is graphed. The first part is linear, enters the window in the second quadrant, goes down and right, crosses the x-axis at approximately x = −0.33, crosses the y-axis at y = −0.25, and ends at the open point (1, −1). The second part is the point (1, 1). The third part is linear, begins at the open point (1, −1), goes up and right, crosses the x-axis at x = 2, and exits the window in the first quadrant. Determine whether f is continuous on its domain. continuous not continuous If it is not continuous on its domain, say why. lim x→1+ f(x) ≠ lim x→1− f(x), so lim x→1 f(x) does not exist. The function is not defined at x = 1. The graph is continuous on its domain. lim x→1 f(x) = −1 ≠ f(1)

The graph of a function f is given. The x y-coordinate plane is given. A curve with 2 parts is graphed. The first part is linear, enters the window in the second quadrant, goes down and right, crosses the x-axis at x = −1, and ends at the open point (0, −1). The second part is linear, begins at the closed point (0, 2), goes down and right, crosses the x-axis at x = 2, and exits the window in the fourth quadrant. Determine whether f is continuous on its domain. continuous not continuous If it is not continuous on its domain, say why. lim x→0 f(x) ≠ f(0) lim x→0+ f(x) ≠ lim x→0− f(x), so lim x→0 f(x) does not exist. The graph has discontinuities at the end points. The graph is continuous on its domain.

Chapter 9 Solutions

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All the points for relative extrema and identify any saddle point for the given function.

Solution Summary: The author explains that the given function has a relative minimum at (x,y).

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To find: All the points for relative extrema and identify any saddle point for the given function.

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Chapter 9 Solutions

To find:

All the points for relative extrema and identify any saddle point for the given function.