Find the linear approximation of the function f(x) = √16-x at a = 0 and use it to approximate the numbers √15.9 and √15.99.

L(x) = f(a) + f'(a)(x - a); where f(x) = sqrt(16 - x) and a = 0.
f'(x) = -1/(2sqrt(16 - x)),
f'(0) = -1/(2sqrt(16 - 0)) = -1/(2sqrt(16)) = -1/2(4) = -1/8,
f(0) = sqrt(16 - 0) = sqrt(16) = 4

L(x) = 4 - 1/8(x - 0) = 4 - 1/8x.

sqrt(15.9) = sqrt(16 - 0.1) = 4 - 1/8(0.1) = 4 - 0.0125 = 3.9875

sqrt(15.99) = sqrt(16 - 0.01) = 4 - 1/8(0.01) = 4 - 0.00125 = 3.99875

Yes, two vertical lines can be complementary