Approximating the value of $\frac{1}{\sqrt{1.1}}$ using Linear approximation of $\frac{1}{\sqrt{1+x}}$. How do I calculate approximately the value of $\frac{1}{\sqrt{1.1}}$ with Linear approximation of the function $\frac{1}{\sqrt{1+x}}$ around point $0$.
And here is a follow-up question: Show that the error in the approximation of $\frac{1}{\sqrt{1.1}}$ in previous question is between  $0.001$ and  $0.01$
Here is what I have got after trying to find the Mclaurin expansion:
$$ f(x) = 1 - \frac{x}{2}  +\frac{3x^2}{8} + R_n(x)$$
im not sure whether if the derivatives were correct, but how do I continue from here? can someone give a clear explanation? 


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 A: You have written a quadratic expansion, and your derivatives are not correct. 
The linear expansion of $f(x) = \dfrac{1}{\sqrt{1 + x}}$ is just 
$$
f(x) \sim f(0) + f'(0)x
$$
so
$$
f(0.1) \sim 1 -\dfrac{1}{2}0.1 = 0.95 
$$
The error can be estimated by
$$
E \leq \dfrac{M}{2!}|0.1|^2
$$
where 
$$
M \geq |f''(x)| \,\, \forall x \in (0.9,1.1)
$$
is an upper bound for the second derivative.
Compute
$$
f''(x) = \left( -\dfrac12 (1 + x)^{-3/2} \right)' = \dfrac{3}{4} (1 + x)^{-5/2} = \dfrac{3}{4} \dfrac{1}{(1 + x)^{5/2}};
$$
this gets larger as the denominator gets smaller, so plugging in $x = 0.9$ gives you an upper bound. 
Can you take it from here?
A: $$(1 + x )^{-1/2} = 1^{-1/2} - \frac12 1^{-3/2}h + \cdots =
1-\frac h{ 2} + \cdots $$ now put $h = 0.1$ to find the linear approximation $$ \frac1{\sqrt{1.1}} \simeq 1-\frac {0.1}{2} = 0.95$$
$\bf edit:$  estimating the error in the approximation.
$\begin{align}\sqrt{1.1} - 1 &= \int_1^{1.1} d x^{-1/2}\\
& = \int_1^{1.1} -\frac12 x^{-3/2}\, dx \\
&= -\frac12 x^{-3/2}(x-1.1)|1^{1.1} + \int_1^{1.1}\frac34(1.1-x)x^{-5/2}\, dx\\
&= -\frac{0.1}2  + \frac34\int_1^{1.1}(1.1-x)x^{-5/2}\, dx\\
&=-\frac{0.1}2  + \frac38 (0.1)^2M, \text{ where } \frac1{{1.1}^{5/2}} < M < 1\\
\end{align}$
taking the worst $$ 0< \sqrt{1.1} - 0.95 < 0.00375. $$
i hope i  did not make any serious errors.
