Question

How do you find the square root of 79 using the linearization techniques?

Answer 1

Linearization means using a tangent line to evaluate the value of an expression. We can approximate a function at a particular point to find the value of the function. To do this we 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 answer:
We are asked to find the value of the square root of 79. So let’s say the function is $\sqrt{x}$, we need to find the equation of the tangent at point $(81,9)$, as this point on the graph of the function has the Y-coordinate which is closest to the 79.
The equation of the tangent is, $y-f(a)=m(x-a)$. We are substituting $a=81$.
$\Rightarrow y-\sqrt{81}={{\left. \dfrac{d\sqrt{x}}{dx} \right|}_{x=81}}\left( x-81 \right)$
We know that the derivative of $\sqrt{x}$ is $\dfrac{1}{2\sqrt{x}}$. Substituting is above, we get

& \Rightarrow y-9={{\left. \dfrac{1}{2\sqrt{x}} \right|}_{x=81}}\left( x-81 \right) \\\ & \Rightarrow y-9=\dfrac{1}{18}\left( x-81 \right) \\\ \end{aligned}$$ We want to find the functional value at 79, substituting $$x=79$$ in the above equation, we get $$\Rightarrow y-9=\dfrac{1}{18}\left( 79-81 \right)$$ Simplifying the right-hand side of the above equation, we get $$\Rightarrow y-9=-0.1111$$ Adding 9 to both sides of the above equation, we get $$\Rightarrow y=9-0.1111=8.9999$$ Hence, the square root of 79 by linearization is $$8.9999$$. **Note:** To solve problems based on linearization we should know which function to use to find the tangent equation. The slope at the point which we chose should not be undefined. Also, the function should not become undefined at the given point. We need to keep these things in mind while choosing the function.