derivatives - Application of Mean Value Theorem and Interval

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!

Answer

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 $$\frac{1}{8}<\frac{\sqrt{53}-7}{2}<\frac{1}{7}$$
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)