Question

How do you use the linear approximation to the square root function to estimate the square root $\sqrt {8.95}$ ?

Answer 1

Hint : In order to determine the approximate value of the square root $\sqrt {8.95}$ , use the technique of linearisation of the function $f\left( x \right) = \sqrt x$ . We can approximate a function at a particular point to find the value of the function. To do this we first have to find the equation of the tangent line at that point. We can find the equation of the tangent using the coordinates and the derivative of the function at that point.

Complete step by step solution:
We are given a square root in the form $\sqrt {8.95}$ .
If we generalise the square root , we have the function of the square root as $f\left( x \right) = \sqrt x$ .
So to find the square root of $8.95$ , we will be using the linearisation technique. As the function is $\sqrt x$ , we first need to find out the equation of the tangent at point $\left( {9,3} \right)$ , as this point on the graph of the function has the y-coordinate which is actually closest to the values $8.95$ .
So, the equation of the tangent is $y - f\left( a \right) = m\left( {x - a} \right)$ where $m$ is the slope of the tangent.
Slope at point $x = 9$ will the derivative of the function at point $x = 9$ .So, $m = \dfrac{d}{{dx}}\left( {\sqrt x } \right){|_{x = 9}}$ .
Now Rewriting the equation of tangent by putting all the values , we have
$\Rightarrow y - f\left( 9 \right) = \dfrac{d}{{dx}}\left( {\sqrt x } \right){|_{x = 9}}\left( {x - 9} \right)$
As we know the derivative of $\dfrac{d}{{dx}}\sqrt x = \dfrac{1}{{2\sqrt x }}$ and $f\left( 9 \right) = \sqrt 9 = 3$ . Substituting these values in above ,we get
$\Rightarrow y - 3 = \dfrac{1}{{2\sqrt x }}{|_{x = 9}}\left( {x - 9} \right) \\\ \Rightarrow y - 3 = \dfrac{1}{{2\sqrt 9 }}\left( {x - 9} \right) \\\ \Rightarrow y - 3 = \dfrac{1}{{2\left( 3 \right)}}\left( {x - 9} \right) \\\ \Rightarrow y - 3 = \dfrac{1}{6}\left( {x - 9} \right) \;$
Since, we want the functional value at $8.95$ , so substituting $x = 8.95$ in the above equation , we have
$\Rightarrow y - 3 = \dfrac{1}{6}\left( {8.95 - 9} \right) \\\ \Rightarrow y - 3 = \dfrac{1}{6}\left( { - 0.05} \right) \;$
Solving the above expression for $y$ , we have
$\Rightarrow y = 3 - 0.008\overline 3 \\\ \Rightarrow y = 2.991\overline 6 \;$
$\therefore y = f\left( {8.95} \right) = 2.991\overline 6$
Therefore the approximate value for the square root $\sqrt {8.95}$ is equal to $2.991\overline 6$
So, the correct answer is “ $2.991\overline 6$ ”.

Note : 1. Linearization means using a tangent line to evaluate the value of an expression.
2.To solve such problems based on linearisation, we should always know which function to use to find the equation of tangent.
3.Remember, the slope at the point, which we have chosen should always be defined.
4. The function should not become undefined at the given point.