An example with negative dx
When using linear approximations, x doesn't have to be bigger
than a.
Here is an example where x is slightly less than a. The
work is the same as when x is larger than a:
Example: If ln(10)=2.30258 (to 5 decimal places),
what is ln(9.95)?
Solution: In this problem we're working with f(x)=ln(x), a=10, and x=9.95, so x−a=−0.05.
Since f′(x)=ddx(ln(x))=1x,
we have f′(a)=110.
Our linearization is then:
L(x)=2.30258+110(x−10).
Plugging in x=9.95 gives
ln(9.95)≈L(9.95)=2.30258+110(−0.05)=2.29758
In fact, ln(9.95)=2.29757 to 5 decimal places, so we were off
by 0.00001. The linear approximation isn't exact, but it is very,
very good.
Here is the same calculation in terms of differentials:
dx=9.95−10=−0.05df=dx10=−0.005f(x)≈f(a)+df=2.30258−0.005=2.29758
|