Answer:
See below in bold.
Step-by-step explanation:
f(x, y, z) = x^2 + y^2 + z^2 at (6, 6, 7)
Taking partial derivatives
f'x = 2x + y^2 + z^2 --> f'x(6,6,7) = 12+36+49= 97
f'y = x^2 + 2y + z^2---> f'y(6,6,7) = 36 + 12 + 49 = 97
f'z = x^2 + y^2 + 2z ----> f'z(6,6,7) = 36 + 36 + 14 = 86.
f(6,6,7) = 36 + 36 + 49 = 121
So our linear approximation is:
L(x,y,z) = 121 + 97(x - 6) + 97(y - 6) + 86(z - 7)
Substituting the given values
f(6.012, 5.972,6.982)
= L(6.012, 5.972, 6.982) ≈ 121 + 97(6.012 - 6) + 97(5.972 - 6) + 86(6.982 - 7)
= 117.9.
