Question

# Use the table of values of f(x, y) to estimate the values of fx(3, 2), fx(3, 2.2), and fxy(3, 2). \begin{array}{|c|c|}\hline y & 1.8 & 2.0 & 2.2 \\ \hline x & & & \\ \hline 2.5 & 12.5 & 10.2 & 9.3 \\ \hline 3.0 & 18.1 & 17.5 & 15.9 \\ \hline 3.5 & 20.0 & 22.4 & 26.1 \\ \hline \end{array}

Use the table of values of $$f(x, y)$$ to estimate the values of $$fx(3, 2)$$, $$fx(3, 2.2)$$, and $$fxy(3, 2)$$.
$$\begin{array}{|c|c|}\hline y & 1.8 & 2.0 & 2.2 \\ \hline x & & & \\ \hline 2.5 & 12.5 & 10.2 & 9.3 \\ \hline 3.0 & 18.1 & 17.5 & 15.9 \\ \hline 3.5 & 20.0 & 22.4 & 26.1 \\ \hline \end{array}$$

2021-06-10

Step 1
$$fx(3.2)=\frac{(f(3.5,\ 2)-f(2.5,\ 5))}{(3.5-2.5)}$$
$$=\frac{(22.4-10.2)}{1.0}$$
$$=\frac{12.2}{1.0}$$
$$=12.2$$
$$fx(3,\ 2.2)=\frac{(f(3.5,\ 2.2)-f(2.5,\ 2.2))}{(3.5-2.5)}$$
$$=\frac{26.1-9.3)}{1}=16.8$$
For $$fxy(3, 2)$$, we have choices. If we don't care about centering the y, we can use the results already gathered here.
Then, $$fxy(3.2)=\frac{(fx(3,\ 2.2)-fx(3.2))}{(2.2-2)}$$
$$=\frac{(16.8-12.2)}{2}=23$$
If we had wanted to center it, we could have used
$$\frac{(f(3.5,\ 2.2)-f(3.5,\ 1.8)-f(2.5,\ 2.2)+f(2.5,\ 1.8))}{((2.2-1.8)\cdot(3.5-2.5)}$$
$$=\frac{(26.1-20-9.3+12.5)}{(4\cdot1)}=23.25$$