Final answer:
The linearization L(x, y) of the function f(x, y) = x^2 + y^2 + 1 at the point (1,1) is L(x, y) = 2x + 2y - 1.
Step-by-step explanation:
To find the linearization L(x, y) of the function f(x, y) = x^2 + y^2 + 1, we need to evaluate the function and its partial derivatives at the given points (1,1) and (0,0).
Let's start with the point (1,1). The partial derivatives of f with respect to x and y are:
- fx(x, y) = 2x
- fy(x, y) = 2y
Evaluating the partial derivatives at (1,1) gives:
- fx(1, 1) = 2(1) = 2
- fy(1, 1) = 2(1) = 2
The function value at (1,1) is f(1, 1) = 1^2 + 1^2 + 1 = 3. The linearization L(x, y) at point (1,1) will be:
L(x, y) = f(a, b) + fx(a, b)(x - a) + fy(a, b)(y - b)
Substituting the values in we get:
L(x, y) = 3 + 2(x - 1) + 2(y - 1)
Simplifying, the equation of the tangent plane is:
L(x, y) = 2x + 2y - 1