To find the linear approximation of the function f(x, y, z) = x² + y² + z² at the point (3, 2, 6), we use the formula for the linear approximation, which is:
L(x, y, z) = f(a, b, c) + fx(a, b, c)(x – a) + fy(a, b, c)(y – b) + fz(a, b, c)(z – c)
Here, (a, b, c) = (3, 2, 6).
First, we need to evaluate the function at the point (3, 2, 6):
f(3, 2, 6) = 3² + 2² + 6² = 9 + 4 + 36 = 49
Next, we compute the partial derivatives:
- fx(x, y, z) = 2x
- fy(x, y, z) = 2y
- fz(x, y, z) = 2z
Now we evaluate these derivatives at the point (3, 2, 6):
- fx(3, 2, 6) = 2(3) = 6
- fy(3, 2, 6) = 2(2) = 4
- fz(3, 2, 6) = 2(6) = 12
Now, we substitute everything into the linear approximation formula:
L(x, y, z) = 49 + 6(x – 3) + 4(y – 2) + 12(z – 6)
Expanding this, we get:
L(x, y, z) = 49 + 6x – 18 + 4y – 8 + 12z – 72
L(x, y, z) = 6x + 4y + 12z – 49
Therefore, the linear approximation of the function f(x, y, z) at the point (3, 2, 6) is:
L(x, y, z) = 6x + 4y + 12z – 49.