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

and

how should we take $f(x)$?

Thanks in advance.

Regards

1 Answers 1

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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 $a$?2011-08-18
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    keep in mind $802011-08-18
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    Oh, so what the problem ends there? Is it only adds $80 anything else?2011-08-18
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    @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?2011-08-18
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    I guess another clue is that evaluating $1/(2\sqrt{a})$ at anything other than $81$ is going to be pretty wild.2011-08-18
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    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}.$2011-08-18
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    excelent!! thanks so much =)2011-08-18