Find the linearization $L(x,y)$ of the function at each point.
$$f(x,y)=x^2+y^2+1$$
a. $(4,1)$
b. $(0,3)$
a. $L(x,y)=$

Step 1 ：Given the function $f(x,y)=x^2+y^2+1$, we want to find its linearization at the point (4,1).

Step 2 ：The linearization of a function at a point $(a,b)$ is given by the formula: $$L(x,y)=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)$$ where $f_x$ and $f_y$ are the partial derivatives of $f$ with respect to $x$ and $y$ respectively.

Step 3 ：First, we need to find the partial derivatives $f_x$ and $f_y$. For the given function, $f_x=2x$ and $f_y=2y$.

Step 4 ：Substitute the point $(4,1)$ into these derivatives to find $f_x(4,1)$ and $f_y(4,1)$. We get $f_x(4,1)=8$ and $f_y(4,1)=2$.

Step 5 ：Substitute these values, along with the point $(4,1)$, into the formula for the linearization. We get $$L(x,y)=f(4,1)+8(x-4)+2(y-1)$$.

Step 6 ：Simplify the above expression to get the final linearization of the function at the point $(4,1)$.

Step 7 ：$\boxed{{\displaystyle L(x,y)=8x+2y-16}}$ is the final answer.