Why is $f(x)$ here not changing as 12 times as x is chanigng? Here we have $f(x)=x^{3}$, if we substitute x with 2, we will get $f(2)=8$. Now if we took the derivative of $f(x)=x^{3}$, we will get $f'(x)=3x^{2}$, and $f'(2)=12$. We can tell from this that a change in x, will change $f(x)$ 12 times as fast as x is changing. 
$f(2.001)=8.012006001$
$f(2.002)=8.024024008$
$f(2.003)=8.036054027$
I can see the pattern here, it is changing 12 times as x is changing, however why is it when I try with for example 2.1 I don't get the same pattern? 
$f(2.1)=9.261$
$f(2.2)=10.648$
$f(2.3)=12.167$
 A: You actually do get the same pattern, it's just that the numbers in the pattern have all started to flow together. $$f(2.001) = 8 + .001 * 12 + 6 * (.001)^2 + (.001)^3$$ $$f(2.002) = 8 + .002 * 12 + 6 * (.002)^2 + (.002)^3$$ $$...$$ $$f(2.1) = 8 + .1 * 12 + 6 * (.1)^2 + (.1)^3 = 9.261$$
A: $9.261 - 8 = 1.261 = 12.61*0.1 \approx 12*0.1$
$10.648 - 9.261 = 1.387=13.87*0.1 \approx 12*0.1$
So you do get the same pattern.
....
$f(x)$ is increasing $12$ times faster than $x$ only at exactly $x=2$.  At just a tiny smidgen past $x=2$ it will be increasing at a faster rate then $12$. 
So the distance travel from $x=2$ to $x=2 + a$ will have to be more than $12a$ because $f(x)$ would be traveling faster then $12$ for all the time $x \in (2, 2+a]$.
On the other hand, at $x=2+a$ we have $f(x)$ is increasing at $3(4+4a + a^2)=12 +12a + 3a^2$.  wheres for all $x \in [2,2+a)$, $f(x)$ was increasing at rate slower than $12 + 12a + 3a^2$.  So the distance travelled will be less than $a(12+12a+3a^2)$.
So the total distance $d$ will be $12a < d < 12a+12a^2+3a^2$. 
For very small values of $a$ (particularly in comparison to $x$) the margin of error is small but for large values it is not.
$f(2) = 8; f(2.01)= 8.120601$
$f'(2) = 12; f'(2.01) = 12.1203$
So $f(2) + 12*.01 < f(2.01) < f(2) + 12.1203*0.01\implies 8.12 < 8.120601 < 8.121203$.
But $f(2.1) = 9.261$ and $f'(2.1) = 3(2.1)^2 = 13.21$ so 
$f(2) + 12*.1 < f(2.1) < f(2) + 13.21*1 \implies 9.2 < 9.261 < 9.321$.
All is as expected.
A: Linear approximations are just  tangent line approximations and they are local. 
That is they work the  best if your change in $x$ is very small. 
In your case, your change of $0.001$ is good but the change of $0.1$ is not very good. 
As you noticed, for  $2.2$ and $2.3$ the estimates are even worse.
For $2.3$ your estimate is $8+12(.3)=11.6$ while the exact value is $2.3 ^3 = 12.2$ an error of $.6$ which is rather large.
