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 $L(x) = f(a) + f'(a)(x - a)$. 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'\left( a \right) = \dfrac{d}{{da}}{\left( a \right)^{\dfrac{1}{2}}}$
 As we already know, $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$ .
Apply the formula in the above step.
$ \Rightarrow f'\left( a \right) = \dfrac{1}{2}{a^{\dfrac{1}{2} - 1}}$
That is equal to,
$ \Rightarrow f'\left( a \right) = \dfrac{1}{2}{a^{ - \dfrac{1}{2}}}$
We can also write the above step as,
$ \Rightarrow f'\left( a \right) = \dfrac{1}{{2\sqrt a }}$
Here, the value of ‘a’ is 4. Put the value of ‘a’ in the above equation.
$ \Rightarrow f'\left( 4 \right) = \dfrac{1}{{2\sqrt 4 }}$
The square root of 4 is 2.
Therefore,
$ \Rightarrow f'\left( 4 \right) = \dfrac{1}{4}$
Now, let us find the approximate value.
$ \Rightarrow L(x) = f(a) + f'(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'(4)(4.400 - 4)$
Now, put $f\left( 4 \right) = 2$and $f'\left( 4 \right) = \dfrac{1}{4}$ in the above equation.
$ \Rightarrow L(4.400) = 2 + \dfrac{1}{4}(0.400)$
That is equal to,
 $ \Rightarrow L(4.400) = 2 + \dfrac{1}{4}\left( {\dfrac{4}{{10}}} \right)$
Simplify the right-hand side. We will get,
$ \Rightarrow L(4.400) = 2 + \dfrac{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.