How do you estimate the quantity using the linear approximation of ${{\left( 3.9 \right)}^{\dfrac{1}{2}}}$?

Hint: We will use the linear approximation formula to find the linear approximation of the given square root number. The linear approximation formula is given by $L\left( x \right)=f\left( a \right)+f'\left( a \right)\left( x-a \right)$ where a is the value we need to consider close to the given value.

Complete step-by-step solution:
We have been given a quantity ${{\left( 3.9 \right)}^{\dfrac{1}{2}}}$.
We have to estimate the given quantity using the linear approximation.
We know that the linear approximation of a function is given as $L\left( x \right)=f\left( a \right)+f'\left( a \right)\left( x-a \right)$.
Here in this question we have to estimate the function $f\left( 3.9 \right)=\sqrt{3.9}$
Now, first we need to choose a value closer to the given number which is the value of $a$.
Let us consider $a=4$ so we will get
$\begin{align}
  & \Rightarrow f\left( a \right)=f\left( 4 \right) \\
 & \Rightarrow f\left( 4 \right)=\sqrt{4} \\
 & \Rightarrow f\left( 4 \right)=2 \\
\end{align}$
Now, we know that if $f\left( x \right)=\sqrt{x}$ then $f'\left( x \right)=\dfrac{1}{2\sqrt{x}}$
So we get $f'\left( 4 \right)=\dfrac{1}{2\sqrt{4}}$
Now, substituting the values in the formula of linear approximation we will get
$\Rightarrow L\left( 3.9 \right)=2+\dfrac{1}{2\sqrt{4}}\left( 3.9-4 \right)$
Simplifying the above obtained equation we will get
$\begin{align}
  & \Rightarrow L\left( 3.9 \right)=2+\dfrac{1}{2\times 2}\left( -0.1 \right) \\
 & \Rightarrow L\left( 3.9 \right)=2+\dfrac{1}{4}\left( -0.1 \right) \\
 & \Rightarrow L\left( 3.9 \right)=2+0.25\left( -0.1 \right) \\
 & \Rightarrow L\left( 3.9 \right)=2-0.025 \\
 & \Rightarrow L\left( 3.9 \right)=1.975 \\
\end{align}$
Hence we get ${{\left( 3.9 \right)}^{\dfrac{1}{2}}}\approx 1.975$

Note: The point to be noted is that always choose the value of $a$ closer to the given number such as the square root of that number can be easily calculated. The method of approximation is useful to find the value of a function at any point. We can easily find the value of a function at any particular point.