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Use the Mean Value Theorem to estimate the value of $\sqrt{80}$.


how should we take $f(x)$?

Thanks in advance.


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up vote 6 down vote accepted

You want to estimate a value of $f(x) = \sqrt{x}$, so that's a decent place to start. The mean value theorem says that there's an $a \in (80, 81)$ such that $$ \frac{f(81) - f(80)}{81 - 80} = f'(a). $$ I don't know what $a$ is, but you know $f(81)$ and you hopefully know how to write down $f'$. How small can $f'(a)$ be?

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Thanks. When I do that, I get: $$\sqrt {80} = 9 - \frac{1}{{2\sqrt a }}$$ So what do I do next? Would you give values ​​to the number $a$? – mathsalomon Aug 18 '11 at 5:54
keep in mind $80<a<81$ – marwalix Aug 18 '11 at 6:15
Oh, so what the problem ends there? Is it only adds $80<a<81$ anything else? – mathsalomon Aug 18 '11 at 6:32
@mathsalomon I waffled a bit on that point. My feeling is that this is a very optimistic estimate: get an upper bound on what $\sqrt{80}$ could be. It turns out to be a decent approximation (to within three decimal places) of the actual value. I don't see anything else to grab onto -- do you? – Dylan Moreland Aug 18 '11 at 6:39
You could make a very crude estimate and say that $\sqrt{a} >8,$ which would (using also $\sqrt{a} <9$) that $$9 - \frac{1}{16} < \sqrt{80} < 9 - \frac{1}{18}.$$ Or a more informed original guess might have been that $\sqrt{a} > 8.5,$ giving $$ 9 - \frac{1}{17} < \sqrt{80} < 9 - \frac{1}{18},$$ placing $\sqrt{80}$ in an interval of length $\frac{1}{306}.$ – Geoff Robinson Aug 18 '11 at 8:27

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