# Estimate value using Lagrange's MVT

Estimate the value of $51^{1/2}$ using Lagrange's MVT. Answer both in terms of inequalities and approximately estimated value.

My method: Let $f(x)=x^{1/2}$ defined in $[49,51]$ and $f'(x)=\frac{x^{-3/2}}{2}$ which also exits in $(49,51)$

Applying Lagrange's MVT, let $c$ belong to $(49,51)$

We have $f'(c)=\frac{f(51)-f(49)}{2}$ and finally $f(51)=7+c^{-3/2}$

I am stuck at this step, how do I get an inequality as well as the value of $f(51)$?

$f(x) = \sqrt{x}, \ f'(x) = \frac{1}{2\sqrt{x}}$

Using the MVT on the interval $[49,51]$, $f(51) = f(49) + 2f'(c)$, where $c\in [49,51]$. Because $f(x)$ is a strictly increasing function, $f(51) > f(49)$. Also because $f'(x)$ is decreasing it has a maximum value at $49$ (on the interval $[49,51]$). We use $49$ as the starting point of the interval because it has a natural square root, $7$.

\begin{align} 7 < f(51) = 7 + 2f'(c) &\leq 7 + 2f'(49) = 7 + \frac{1}{\sqrt{49}} = 7 + \frac{1}{7}\\ 7 < \sqrt{51} &\leq 7.142857142.. \end{align}

A linear aproximation of a function is an estimate of the function value at $f(x_0 + \Delta x)$ given the value at $f(x_0)$, and first derivative $f'(x_0)$, then

$$f(x_0 + \Delta x) \approx f(x_0) + f'(x_0)\Delta x$$ This has a nice geometric interpretation. For your example the linear approximation of $f(51)$ is

$$f(49 + 2) = f(49) + 2f'(49) = 7 + \frac{1}{7} = 7.142857142..$$

Same as the upper bound of the value by the MVT.

The actual value is $$\sqrt{51} = 7.141428..$$

So you can see that for small $\Delta x$ a linear approximation can be pretty accurate.