(a) Check the local linearity of $f(x,y) = e^{-x}\cos\!\left(y\right)$ near $x=1,\ y=1.5$ by filling in the following table of values of $f$ for $x=0.9,\ 1,\ 1.1$ and $y=1.4,\ 1.5,\ 1.6$. Express values of $f$ with 4 digits after the decimal point.

 $x =$ 0.9 1 1.1 $y = 1.4$ $y = 1.5$ $y = 1.6$

(b) Next, fill in the table for the values $x=0.99,\ 1,\ 1.01$ and $y = 1.49,\ 1.5,\ 1.51,$ again showing 4 digits after the decimal point.

 $x =$ 0.99 1 1.01 $y = 1.49$ $y = 1.5$ $y = 1.51$

Notice if the two tables look nearly linear, and whether the second looks more linear than the first (in particular, think about how you would decide if they were linear, or if the one were more closely linear than the other).

(c) Give the local linearization of $f(x,y) = e^{-x}\cos\!\left(y\right)$ at $(1,1.5)$:
Using the second of your tables:
$f(x,y) \approx$
Using the fact that $f_x(x,y) = -e^{-x}\cos\!\left(y\right)$ and $f_y(x,y) = -e^{-x}\sin\!\left(y\right)$:
$f(x,y) \approx$