# Calculate the 3rd order Taylor polynomial about $x=1$?

Calculate the $3$rd order Taylor polynomial about $x=1$ for the function $f:[-3:\infty) \longrightarrow \mathbb{R}$ given by $f(x)=\sqrt{x+3}$.

I know that the formula for the Taylor Polynomial is:

$$f(x) = \sum_{n=0}^{\infty}\frac{f^{(n)}(x_0)}{n!}(x - x_0)^n$$

However the restricted domain $[-3, \infty)$ confuses me. Does this mean that I start with $n=-3$? In which case how would I calculate $f^{-3}(x)$? Would it be the integral?

The domain is restricted because $\sqrt{x+3}$ is not a real number for $x<-3$. This affects the interval of convergence of the Taylor series (obviously it will not converge for $x<-3$, because there is nothing for it to converge to), but not the calculation of the Taylor series, which just depends on taking derivatives at (in this case) $x=1$. You don't have to do anything different from usual in calculating it.