Solve for: sqrt(2)

Expression: $\sqrt{ 2 }$

To approximate the root using linearization, define a function $f\left( x \right)=\sqrt{ x }$

$f\left( x \right)=\sqrt{ x }$

Find the center of approximation and write it next to the function

$\begin{array} { l }f\left( x \right)=\sqrt{ x },& a=1\end{array}$

To find the linearization of the function, find the value of the function and its derivative at the center of approximation $a=1$

$\begin{array} { l }\begin{array} { l }f\left( x \right)=\sqrt{ x },& a=1,& f\left( a \right)=?\end{array},\\\begin{array} { l }f\left( x \right)=\sqrt{ x },& a=1,& f '\left( a \right)=?\end{array}\end{array}$

Calculate the function value for $a=1$

$\begin{array} { l }f\left( a \right)=1,\\\begin{array} { l }f\left( x \right)=\sqrt{ x },& a=1,& f '\left( a \right)=?\end{array}\end{array}$

Find the value of the derivative for $a=1$

$\begin{array} { l }f\left( a \right)=1,\\f '\left( a \right)=\frac{ 1 }{ 2 }\end{array}$

The linearization of $f\left( x \right)=\sqrt{ x }$ centered at $1$ is $L\left( x \right)=f\left( a \right)+f '\left( a \right)\left( x-a \right)$ where $a=1$, $f\left( a \right)=1$ and $f '\left( a \right)=\frac{ 1 }{ 2 }$

$L\left( x \right)=1+\frac{ 1 }{ 2 } \times \left( x-1 \right)$

Use the linearization to approximate the root $\sqrt{ 2 }$ by substituting $2$ for $x$

$L\left( 2 \right)=1+\frac{ 1 }{ 2 } \times \left( 2-1 \right)$

Simplify the expression

$L\left( 2 \right)=\frac{ 3 }{ 2 }$

Convert the fraction into a decimal

$L\left( 2 \right)=1.5$

The approximation of $\sqrt{ 2 }$ using linearization at $1$ is $1.5$