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