# 5 let f x y z x 2 xy y 2 z 3 4 a find a point x y z

This preview shows page 3. Sign up to view the full content.

phase-out of welfare benefits as income increases has a similar effect. 5. Let f ( x, y, z ) = x 2 + xy + y 2 + z 3 - 4. a ) Find a point ( x 0 , y 0 , z 0 ) satisfying f ( x 0 , y 0 , z 0 ) = 0 . Answer: There are many solutions: The point ( x 0 , y 0 , z 0 ) = (1 , 1 , 1) seemed to be the consensus choice, but (0 , 2 , 0), (2 , 0 , 0), (5 , 1 , - 3), (1 , 5 , - 3), ( - 2 , 1 , 1), (1 , - 2 , 1), and (2 , 2 , 2) are other integral solutions. Of course, you are not restricted to integers, but they often make calculation easier. b ) Can x be expressed as a function g ( y, z ) in some neighborhood of ( x 0 , y 0 , z 0 )? Answer: We use the point ( x 0 , y 0 , z 0 ) = (1 , 1 , 1). Here ∂f/∂x = 2 x 0 + y 0 = 3 6 = 0. The Implicit Function Theorem now implies that we can write x = g ( y, z ) in some neighborhood of (1 , 1 , 1). Note: In this case, we can use the quadratic formula to find an expression for g , g ( y, z ) = - y + p 16 - 3 y 2 - 4 z 3 2 . Of course, this is only valid for 16 - 3 y 2 - 4 z 3 0. Notice that the positive square root is chosen so that g ( y 0 , z 0 ) = g (1 , 1) = 1 = x 0 . Had we used ( - 2 , 1 , 1) as our reference point, we would use the negative square root. c ) Compute dg ( y 0 , z 0 ). Answer: By the Implicit Function Theorem, dg ( y 0 , z 0 ) = - ∂f ∂x - 1 ∂f/∂y ∂f/∂z = - 1 2 x 0 + y 0 x 0 + 2 y 0 3 z 0 = - 1 3 3 3 = - 1 - 1 . Notice that this is an easier computation than using the formula for g given in part ( b ).
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern