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ANSWER ALL PLEASE

9 L+ 4(x 5 - (- -- - +++ a) sketch flx,y) b) Draw the level curves f(x, y) = for k=2,100,-1,-2 c) Compute f (3,3). what point

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(1) Here is the graph ({\color{Red} x},{\color{DarkGreen} y}, {\color{Blue} f(x,y)})
-0 10 1 1 T 2

b.
-5 4. k=2 3 27 k=1 -k=0 k=-1 k=-2 1- 2 3

c. We have f(3,3) = -(-1)^2 + -(0)^2 + 2 =1 which lies in a level curve

-5- -4 k=2 f(3,3) 3 -2 k=1 -k=0 k=-1 k=-2 -1- 2 3

d. Remember that the first derivate says if the function is increasing/decreasing and the second derivate talks about the concavity. Therefore,

\begin{matrix} f_x(3,3) & -\\ f_y(3,3) & 0 \\ f_{xx}(3,3) & -\\ f_{yy}(3,3) & - \end{matrix}.

(2) For the square root term we need x- y^2 \geqslant 0 or x\geqslant y^2 . Also, for the term \frac2{x^2 + y^2} we need that x,y are not zero at the same time. For the third term to be defined we need 9- x^2 > 0 or x^2 <9 . These are the corresponding regions,

1596244832924_desmos-graph4.png

and their intersction is

1596244907834_desmos-graph5.png

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