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Use linear approximation (or differentials) to estimate:

$$\sqrt {99.2}$$

What am I supposed to do with this? I am not given $x$ or $dx$.

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Put braces {} around 99.2, so that the sqrt symbol knows the overline should cover the entire thing. – Hurkyl Oct 27 '12 at 22:53
up vote 5 down vote accepted

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, $$\sqrt{99.2} \approx \sqrt{100} + \dfrac12 \dfrac{(99.2-100)}{\sqrt{100}} = 10 - \dfrac12 \dfrac{0.8}{10} = 10 - 0.04 = 9.96$$

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@Henning Thanks for the fix. – user17762 Oct 27 '12 at 22:58
Thank you so much!!!! It makes sense now! – dsta Oct 27 '12 at 23:16

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

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Recall that $f'(a) \approx \tfrac{{f(x) - f(a)}}{{x - a}}$ so $$f(x) \approx f(a) + f'(a)(x - a).$$ Now define $f(x) = \sqrt x $ and use the fact that $99.2 \approx 100$. We seek $f(100)$, so find $f'(x)$, 100 - 99.2, etc. You can also use differentials which is exactly the same since $f(x) - f(a) \approx dy$ and $x - a \approx dx$.

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