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 $R(\xi) > 10^{-4}$

$f(x) = \sqrt{1 + x}$
$f'(x) = \frac{1}{2\sqrt{1+x}}$
$f(0) = 1$
$f'(0) = 1/2$
$f(x + h) = f(x) + h f'(x) +.....$
$\sqrt{1.1} = f(.1) = f(0 + .1) \approx1 + 0.1 / 2 = 1.05$
you got confused by the fact they are using x + 1 in the function - i made a start for you, with two terms it is getting closer to the answer
do you see the important part for you? $\sqrt{1.1} = \sqrt(1 + .1) = f(.1)$ 
since it is easy to workout f(0), f'(0),f''(0) without any square -roots, you then need to centre it around 0 - i.e. f(0 + .1)
