Question

Approximate the root by using a linearization centered at an appropriate nearby number $\sqrt{101}$.

Answer 1

Calculating the root of $\sqrt{101}$:

The closet perfect square to $101$is $100$.

So we center the linearization at $x=100$.

Now, let

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)=\sqrt{\mathrm{x}} \\ \Rightarrow \mathrm{f}\text{'}\left(\mathrm{x}\right)=\frac{1}{2\sqrt{\mathrm{x}}} \\ \Rightarrow \mathrm{f}\text{'}\left(100\right)=\frac{1}{2\sqrt{100}} \\ \Rightarrow \mathrm{f}\text{'}\left(100\right)=\frac{1}{20} \end{gathered}$$

Also,

$\begin{array}{rcl}f\left(100\right)& =& \sqrt{100}\\ & =& 10\end{array}$

So,

$\mathrm{L}\left(\mathrm{x}\right)=10+\frac{1}{20}(\mathrm{x}-100)$

$\begin{array}{rcl}f\left(101\right)& =& \sqrt{101}\\ L\left(101\right)& \approx & 10+\frac{1}{20}(101-100)\\ & \approx & 10+\frac{1}{20}\\ & \approx & \frac{201}{20}\\ & \approx & 10.05\end{array}$

Hence, the approximate root of $\sqrt{101}$ by linearization is $10.05$.