Question

Suppose that f(x,y) is a smooth function and that its partial derivatives have the values, fx(−7,−2)=−1 and fy(−7,−2)=3. Given that f(−7,−2)=−2, use this information to estimate the value of f(−6,−1). Note this is analogous to finding the tangent line approximation to a function of one variable. In fancy terms, it is the first Taylor approximation.

Answer 1

Answer:

$f(-6,-1) =0$

Step-by-step explanation:

We are given the following in the question:

$f_x(-7,-2)=-1\\f_y(-7,-2)=3\\f(-7,-2)=-2$

We have to estimate the value of f(-6,-1)

By Taylor approximation we have:

$f(x,y) =f(-7,-2)+ f_x(-7,-2)(x+7) + f_y(-7,-2)(y+2)\\f(x,y) = -2 + (-1)(x+7) + 3(y+2)\\f(x,y) = -2-x-7+3y+6\\f(x,y) = -x+3y-3$

Now, putting values, we get,

$f(x,y) = -x+3y-3\\f(-6,-1) = -(-6)+3(-1)-3\\f(-6,-1) =0$

is the required value.