There are three parts to this problem. For (a) and (b), all you are doing is entering in the value of the function at the given points. For instance, my function is $f(x,y)=e^{-y}\cos(x)$ near $x=2.5$ and $y=1$ (some of you may have the same function with different $x$ and $y$-values, etc.). My tables for (a) and (b) end up looking something like this.

Again, there is nothing special going on here. We’re just evaluating the function at these points. The thing that WebWork is trying to get you to realize is that the values in the second table are closer to each other than the values in the first table. The function is indeed locally linear.

OK, so what is (c) asking? What (c) is looking for is an approximation to

\[f(x,y)\approx f(2.5,1)+f_x(2.5,1)(x-2.5)+f_y(2.5,1)(y-1)\]

where what we really need to calculate is an approximation to $f_x(2.5,-1)$ and an approximation to $f_y(2.5,-1)$. The first entry in (c) wants you to calculate these values from the second table above. For instance, in my problem with my numbers, I’d have

\[f_x(2.5,1)\approx \displaystyle\frac{-0.2969 - (-0.2925)}{0.02}=-0.220\]

where I’m approximating the $x$-derivative at point $(2.5,1)$ by using the values of the function given in the table. Using this idea, I find that

\[f(x,y)\approx -0.2947 - 0.220(x-2.5)+0.295(y-1)\]

and this answers the first part of (c). The second part of (c) is looking for an answer that is slightly more precise. Here, we use the actual partial derivatives and we find

\[f(x,y)\approx -0.2947+(-0.220166)(x-2.5)+0.294724(y-1).\]

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