# Given f(x) = sqrt(1-x), how do you write the Taylor series about c = -3 with the first four terms?

The general formula is f(c)+f'(c)(x-c)+(f''(c))/(2!)(x-c)^2+(f'''(c))/(3!)(x-c)^3+\cdots.
Since $f \left(x\right) = \setminus \sqrt{1 - x} = {\left(1 - x\right)}^{\frac{1}{2}}$ we get $f ' \left(x\right) = - \setminus \frac{1}{2} {\left(1 - x\right)}^{- \frac{1}{2}}$, $f ' ' \left(x\right) = - \setminus \frac{1}{4} {\left(1 - x\right)}^{- \frac{3}{2}}$, and $f ' ' ' \left(x\right) = - \setminus \frac{3}{8} {\left(1 - x\right)}^{- \frac{5}{2}}$.
Since $c = - 3$, we get $f \left(c\right) = \setminus \sqrt{4} = 2$, $f ' \left(c\right) = - \setminus \frac{1}{2 \setminus \sqrt{4}} = - \setminus \frac{1}{4}$, $f ' ' \left(c\right) = - \setminus \frac{1}{4 \setminus \cdot {4}^{\frac{3}{2}}} = \setminus \frac{1}{32}$, and $f ' ' ' \left(c\right) = - \setminus \frac{3}{8 \setminus \cdot {4}^{\frac{5}{2}}} = - \setminus \frac{3}{256}$.
Therefore, $\setminus \sqrt{1 - x} = 2 - \setminus \frac{1}{4} \left(x + 3\right) - \setminus \frac{1}{64} {\left(x + 3\right)}^{2} - \setminus \frac{1}{512} {\left(x + 3\right)}^{3} + \setminus \cdots$.