Consider a function f : {R}^2 \to {R}^2 for which f(4, -4) = (3, 3) and f{D}f(4, -4) = (5 4)
(-3 0) <-- this is a 2x2 matrix
The linear approximation of f at the point (4, -4) is (written as a row vector) is L(x,y) = (___,___)
thanks,
I have no idea where to. Since they give me the answer and derivatives and linear approx are meant to extrapolate...
To find the linear approximation of the function f at the point (4, -4), we can use the given derivative matrix f'(4, -4).
The linear approximation of a function f(x, y) at a point (a, b) can be written as:
L(x, y) = f(a, b) + f'(a, b) * (x - a, y - b)
In this case, a = 4 and b = -4, and we are given:
f(4, -4) = (3, 3)
f'(4, -4) = (5 4)
(-3 0)
Using the formula for linear approximation, we can substitute the given values into the equation:
L(x, y) = (3, 3) + (5 4) * (x - 4, y - (-4))
(-3 0)
Notice that (x - 4, y - (-4)) is the vector (x - 4, y + 4).
Multiplying the matrix f'(4, -4) with the vector (x - 4, y + 4), we get:
L(x, y) = (3, 3) + (5*(x - 4) + 4*(y + 4), -3*(x - 4))
Expanding the above equation, we can simplify further:
L(x, y) = (3 + 5x - 20 + 4y + 16, -3x + 12)
Combining like terms, we get the linear approximation function as:
L(x, y) = (5x + 4y - 1, -3x + 12)
Therefore, the linear approximation of f at the point (4, -4) is written as the row vector (5x + 4y - 1, -3x + 12).
If you have any further questions, please, let me know!