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Using the mean value theorem establish the inequality $$7\frac{1}{4}<\sqrt{53}<7\frac{2}{7}$$

This is obviously a true statement but can you help me form the interval and what function I should use to prove this using the mean value theorem? I've only done problems where the interval is given.

Thanks!

user2553807
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    You're probably going to want a polynomial with $\sqrt 53$ as a root, and you show that the function is positive at $51/7$ and negative at $29/4$ (or vice versa). This function might even be $x^2-53=0$. – Ian Coley Dec 09 '13 at 20:57

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You can consider the function $f:[0,\infty)\to \mathbb{R}$ given by $f(x)=\sqrt{x}$ and the interval [49, 53]. By the Men Value Theorem there exists some $c\in (49, 53)$ such that $$ \frac{\sqrt{53}-7}{4}= \frac{1}{2\sqrt{c}} $$ or equivalently, $$ \frac{\sqrt{53}-7}{2}= \frac{1}{\sqrt{c}} $$ Because $49<c<64$, then $\frac{1}{8}<\frac{1}{\sqrt{c}}< \frac{1}{7}$. Using this and the previous equality, we arrive at $$ \frac{1}{8}<\frac{\sqrt{53}-7}{2}<\frac{1}{7} $$ Wich implies that $$ 7+\frac{1}{4}<\sqrt{53}<7+ \frac{2}{7} $$ (and I suppose this is the inequality you wanted to show, not the other one you gave in the problem)

Brandon
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  • This looks like an excellent answer, but it is not clear to me how you arrived at the first major equation. Perhaps you could clarify? – abiessu Dec 09 '13 at 23:14
  • The Mean Value theorem states that given a continuos function $f$ on an interval $[a,b]$ that is differentiable on $(a,b)$, there exists a $c\in (a,b)$ such that $(f(b)-f(a))/(b-a)= f'(c)$. Take $b=53, a=49$ and $f$ as given above and note that $f(49)= \sqrt{49}=7$ and that $53-49=4$. – Brandon Dec 09 '13 at 23:34
  • That's right, I forgot the $b-a$ denominator. Thank you! – abiessu Dec 09 '13 at 23:37