Accuracy of linear approximations It's another day of calculus and I'm having trouble with linear approximations; perhaps you guys can help. I am unsure of how to calculate the accuracy of these approximations; let me give you an example.

Verify the given linear approximation at $a = 0$. Then determine the values of $x$ for which the linear approximation is accurate to within $0.1$.
$$1/(1+2x)^4 \approx 1 – 8x$$

I can verify the linear approximation easily enough, but how do I determine its accuracy? Thanks!
 A: You question is a little ambiguous but I assume you mean find the values of $x$ for which the computed function values differ by less than 0.1.
If you are allowed to use a graphing calculator or something similar, just graph the functions
$$f1(x) = \left| {\frac{1}{{{{(1 + 2x)}^4}}} - (1 - 8x)} \right|$$
and 
$$f2(x) = 0.1$$
and see where they intersect. I get the result
$$- 0.04536 \leqslant x \leqslant 0.05539$$
to five decimal places.
Another way is to use the second derivative but that seems too advanced for Precalculus.
A: One computes
$$f(x):={1\over(1+2x)^4}-(1-8x)=40 x^2\ {1+4x+6x^2+3.2 x^3\over(1+2x)^4}\ ,$$
which is $>0$ for $|x|\ll1$. For such $x$ the fraction $q$ on the right hand side is $\doteq1$. Therefore  in a first round we make sure that $$40x^2<0.1\ .\tag{1}$$ The latter is equivalent with $|x|<0.05$.
This is "for all practical purposes" sufficient, but of course we now should take $q$ into account to be really on the safe side. When $|x|<0.05$ to begin with then
$$q<{1+4|x|+6|x|^2+3.2|x|^3\over(1-2|x|)^4}\leq{1.25\over0.9^4}<2\ .$$
This implies that we have to replace the simple condition $(1)$ by $$2\cdot 40x^2<0.1\ ,$$ and this leads to $|x|<0.035$.
