Question

How do you use a linear approximation to the square root function to estimate square roots $\sqrt{4.400}$?

Answer 1

Hint: In this question, we have to find the square root by using the linear approximation to the square root function. The linear approximation of f at ‘a’ is $L(x)=f(a)+f^{\prime }(a)(x-a)$. And the function we want to approximate is $f\left(x\right)=\sqrt{x}$.

Complete step-by-step answer:
In this question, the function we want to approximate is $f\left(x\right)=\sqrt{x}$.
Here, the value of x is 4.400.
Therefore, we want to estimate $f\left(4.400\right)=\sqrt{4.400}$. So, we need a value of ‘a’ that is close to 4.400 and for which we can easily find $f\left(a\right)$.
In this question, we want a number close to 4.4 whose square root is relatively easy to find. So, we select the value of ‘a’ is 4.
Now,
$\Rightarrow f\left(a\right)=\sqrt{a}$
Here, the value of ‘a’ is 4. Put the value of ‘a’ in the above equation.
$\Rightarrow f\left(4\right)=\sqrt{4}$
The square root of 4 is 2.
Therefore,
$\Rightarrow f\left(4\right)=2$
Now, let us find the derivative of $f\left(a\right)$ with respect to a.
$\Rightarrow f^{\prime }\left(a\right)={\displaystyle \frac{d}{da}}{\left(a\right)}^{{\displaystyle \frac{1}{2}}}$
 As we already know, ${\displaystyle \frac{d}{dx}}{x}^n=n{x}^{n-1}$ .
Apply the formula in the above step.
$\Rightarrow f^{\prime }\left(a\right)={\displaystyle \frac{1}{2}}{a}^{{\displaystyle \frac{1}{2}}-1}$
That is equal to,
$\Rightarrow f^{\prime }\left(a\right)={\displaystyle \frac{1}{2}}{a}^{-{\displaystyle \frac{1}{2}}}$
We can also write the above step as,
$\Rightarrow f^{\prime }\left(a\right)={\displaystyle \frac{1}{2\sqrt{a}}}$
Here, the value of ‘a’ is 4. Put the value of ‘a’ in the above equation.
$\Rightarrow f^{\prime }\left(4\right)={\displaystyle \frac{1}{2\sqrt{4}}}$
The square root of 4 is 2.
Therefore,
$\Rightarrow f^{\prime }\left(4\right)={\displaystyle \frac{1}{4}}$
Now, let us find the approximate value.
$\Rightarrow L(x)=f(a)+f^{\prime }(a)(x-a)$
Here, the value of ‘a’ is equal to 4, and the value of ‘x’ is equal to 4.400.
Let us substitute those values.
$\Rightarrow L(4.400)=f(4)+f^{\prime }(4)(4.400-4)$
Now, put $f\left(4\right)=2$and $f^{\prime }\left(4\right)={\displaystyle \frac{1}{4}}$ in the above equation.
$\Rightarrow L(4.400)=2+{\displaystyle \frac{1}{4}}(0.400)$
That is equal to,
 $\Rightarrow L(4.400)=2+{\displaystyle \frac{1}{4}}\left({\displaystyle \frac{4}{10}}\right)$
Simplify the right-hand side. We will get,
$\Rightarrow L(4.400)=2+{\displaystyle \frac{1}{10}}$
Therefore,
$\Rightarrow L(4.400)=2+0.1$
Let us apply addition.
$\Rightarrow L(4.400)=2.1$

Hence, $\sqrt{4.400}$ is equal to 2.1.

Note:
Linear approximation is also known as linearization. It is a method to approximate the value of a function at a particular point. It is a quick and simple method that estimates a value otherwise it is very difficult to find. And square roots are a great example of this.