TITLE: Use a linear approximation (or differentials) to estimate the given number.
QUESTION [5 upvotes]: Use linear approximation (or differentials) to estimate:
$$\sqrt {99.2}$$
What am I supposed to do with this? I am not given $x$ or $dx$.

REPLY [5 votes]: Ue Taylor series for $\sqrt{x}$ about $x = 100$. The reason to expand the Taylor series about $100$ is that $100$ is the closest square to $99.2$.
$$f(x) = f(100) + f'(100) (x-100) + \text{higher order terms}$$
Hence,
$$\sqrt{99.2} \approx \sqrt{100} + \dfrac12 \dfrac{(99.2-100)}{\sqrt{100}} = 10 - \dfrac12 \dfrac{0.8}{10} = 10 - 0.04 = 9.96$$

REPLY [2 votes]: Hint: Is there any number near $99.2$ whose square root is easy?  That will be your $x$ value.