$f\left( x \right)=\sqrt{ x }$
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$$L\left( 2 \right)=1+\frac{ 1 }{ 2 } \times \left( 2-1 \right)$
Simplify the expression$L\left( 2 \right)=\frac{ 3 }{ 2 }$
Convert the fraction into a decimal$L\left( 2 \right)=1.5$
The approximation of $\sqrt{ 2 }$ using linearization at $1$ is $1.5$$\approx1.5$