Find $\sqrt{1.1}$ using Taylor series of the function $\sqrt{x+1}$ in $x^{}_0 = 1$ with error smaller than $10^{-4}$ I should find $\sqrt{1.1}$ using Taylor series of the function $\sqrt{x+1}$  in $x^{}_0=1$ with error smaller than $10^{-4}$.
The first derivatives are 
$$f'(x)=\frac{1}{2\sqrt{x+1}}$$
$$f''(x)=\frac{-1}{4\sqrt{x+1}^ 3}$$
$$f'''(x)=\frac{3}{8\sqrt{x+1}^5}$$
Applying $x^{}_0$ we have:
$$f(1)=\sqrt{2}$$
$$f'(1)=\frac{1}{2\sqrt{2}}$$
$$f''(1)=\frac{-1}{4\sqrt{2}^ 3}$$
$$f'''(1)=\frac{3}{8\sqrt{2}^5}$$
And we can build the Taylor polynomial 
$$T(x)=\sqrt2 + \frac{1}{2\sqrt{2}}(x+1)+\frac{-1}{2!·4\sqrt{2}^3}(x+1)^2+\frac{3}{3!·8\sqrt{2}^5}(x+1)^3+R(\xi)$$
Is everything right until here?
What I don't understand is how can I check that $R(\xi) > 10^{-4}$

Hint:$$f(x_0+h)=f(x_0)+f'(x_0)h+f''(x_0).\dfrac{h^2}{2!}+f^3(x_0).\dfrac{h^3}{3!}+f^4(x_0).\dfrac{h^4}{4!}+o(h^5)\\h=0.1 \\
\to f(x_0+h)=f(x_0)+f'(x_0)(0.1)+f''(x_0).\dfrac{(0.1)^2}{2!}+f^3(x_0).\dfrac{(0.1)^3}{3!}+f^4(x_0).\dfrac{(0.1)^4}{4!}+o((0.1)^5)$$ so in your case $f^4(x_0).h^4$ is in order of $10^{-4}$
and the next terms are in order of $10^{-5},10^{-6},10^{-6},...$ so , it is obvious to do it before term $f^5(x_0).\dfrac{h^5}{5!}=f^5(1).\dfrac{(10^{-5})}{5!}$
