calculus
posted by Anonymous .
Use the linear approximation
(1+x)^k=1+kx to find an approximation for the function f(x)=1/square root of (4+x) for values of x near zero.

Let's go astep further and also work out the quadratic term using the general formula:
(1+x)^k=1+kx + k(k1)/2 x^2 +
k(k1)(k2)/3! x^3 + ....
f(x)= (4 + x)^(1/2) =
4^(1/2) (1 + x/4)^(1/2) =
1/2 [1  x/8 + 3/128 x^2 + ...] =
1/2  x/16 + 3/256 x^2 + ...