# 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).

Figure 1

This problem aims to find the values of a function having alternate independent variables. A table is given to address the values of $x$ and $y$.

These formulas would be required to find the solution:

$f_x(x,y)=\lim_{h \to 0}\dfrac{f(x+h, y)-f(x,y)}{h}$

$f_y(x,y)=\lim_{h\to 0}\dfrac{f(x, y+h)-f(x,y)}{h}$

$f_{xy}=\dfrac{\partial}{\partial y}\left(\frac{\partial f}{\partial x} \right)=\dfrac{\partial}{\partial y}(f_x$

##### Part a:

$f_x(3,2)$ $f_x(x,y)=\lim_{h \to 0}\dfrac{f(x+h, y)-f(x,y)}{h}$ and considering $h=\pm 0.5$

$= \lim_{h \to 0}\dfrac{f(3 \pm 0.5, 2)-f(3,2)}{\pm 0.5}$

Solving for $h=0.5$

$= \dfrac{f(3.5, 2)-f(3,2)}{0.5}$

Using the table to plug in the functions values:

$= \dfrac{22.4-17.5}{0.5}$

$= 9.8$

Now solving for $h=-0.5$

$= \dfrac{f(2.5, 2)-f(3,2)}{-0.5}$

Using the table to plug in the functions values:

$= \dfrac{10.2-17.5}{-0.5}$

$= 14.6$

Taking average of both $\pm 0.5$ answers for the final answer of $f_(3,2)$

$f_x(3,2)=\dfrac{9.8+14.6}{2}$

$f_x(3,2)= 12.2$

##### Part b:

$f_x(3,2.2)$

$f_x(3,2.2)=\lim_{h \to 0}\dfrac{f(3 \pm 0.5, 2.2)-f(3,2.2)}{\pm 0.5}$

Solving for $h=0.5$

$= \dfrac{f(3.5, 2.2)-f(3,2.2)}{0.5}$

Using the table to plug in the functions values:

$= \dfrac{26.1-15.9}{0.5}$

$= 20.4$

Now solving for $h=-0.5$

$= \dfrac{f(2.5, 2.2)-f(3,2.2)}{-0.5}$

Using the table to plug in the functions values:

$=\dfrac{9.3-15.9}{-0.5}$

$=13.2$

Taking average of both $\pm 0.5$ answers for the final answer of $f_(3,2)$

$f_x(3,2.2)=\dfrac{20.4+13.2}{2}$

$f_x(3,2.2) = 16.8$

##### Part c:

$f_xy(3,2)$

$f_{xy}(x,y)=\dfrac{\partial}{\partial y}\left( \frac{\partial f}{\partial x}\right)=\dfrac{\partial}{\partial y} (f_x)$

$=\lim_{h \to 0}\dfrac{f_x(x, y+h)-f_x(x,y)}{h}$

$f_{xy}(3,2)=\lim_{h \to 0}\dfrac{f_x(3, 2+h)-f_x(3,2)}{h}$

Considering $h=\pm 0.2$

Solving for $h=0.2$

$=\dfrac{f_x(3, 2.2)-f_x(3,2)}{0.2}$

Plugging in the answers from part a and part b:

$=\dfrac{16.8-12.2}{0.2}$

$=23$

Now solving for $h=-0.2$

$=\dfrac{f_x(3, 1.8)-f_x(3,2)}{-0.2}$

Solving $f_x(3, 1.8)$ for $h=\pm 0.5$

Solving for $h=0.5$

$f_x(3,1.8)=\lim_{h \to 0}\dfrac{f(3 \pm 0.5, 1.8)-f(3,1.8)}{\pm 0.5}$

$=\dfrac{f(3.5, 1.8)-f(3,1.8)}{0.5}$

Using the table to plug in the functions values:

$=\dfrac{20.0-18.1}{0.5}$

$= 3.8$

Now solving for $h=-0.5$

$= \dfrac{f(2.5, 1.8)-f(3,1.8)}{-0.5}$

Using the table to plug in the functions values:

$= \dfrac{12.5-18.1}{-0.5}$

$= 11.2$

Taking average of $\pm 0.5$ answers for the final answer of $f_x(3,1.8)$

$f_x(3,1.8) = \dfrac{3.8+11.2}{2}$

$f_x(3,1.8) = 7.5$

Substituting $f_x(3,1.8)$ in the main equation above to find $f_{xy}(3,2)$

$f_{xy}(3,2)$ for $h = -2$ becomes:

$= \dfrac{f_x(3, 1.8)-f_x(3,2)}{-0.2}$

Plugging in the values:

$= \dfrac{7.5-12.2}{-0.2}$

$= \dfrac{7.5-12.2}{-0.2}$

$= 23.5$

Taking Average of $h=\pm 0.2$ answers to find the final answer:

$f_{xy}(3,2) = \dfrac{23+23.5}{2}$

$f_{xy}(3,2) = 23.25$

## Numerical Results:

Part a: $f_x(3,2) = 12.2$

Part b: $f_x(3,2.2) = 16.8$

Part c: $f_{xy}(3,2) = 23.25$

## Example

For the given table, find $f_y(2.5, 2)$.

$f_y(x,y) = \lim_{h \to 0} \dfrac{f(x, y+h)-f(x,y)}{h}$

Plugging the values in:

$f_y(2.5,2) = \lim_{h \to 0} \dfrac{f(2.5, 2+h)-f(2.5,2)}{h}$

Solving for $h = \pm 0.2$

For $h = 0.2$

$= \dfrac{f(2.5, 2.2)-f(2.5,2)}{0.2}$

Using the table to plug in the function values:

$= \dfrac{9.3 – 10.2}{0.2}$

$= -4.5$

Now solving for $h=-0.2$

$= \dfrac{f(2.5, 1.8)-f(2.5,2)}{-0.2}$

Using the table to plug in the functions values:

$= \dfrac{12.5-10.2}{-0.2}$

$= – 11.5$

Taking average of $\pm 0.5$ answers for the final answer of $f_y(2.5,2)$:

$f_y(2.5,2) = \dfrac{-4.5-11.5}{2}$

$f_y(2.5,2) = -8$

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