– For the given linear function as $f(x)=\sqrt{4-x}$, calculate the linear approximation at a=0. Based on this linear approximation $L(x)$, approximate the values for given two functions $\sqrt{3.9}$ and $\sqrt{3.99}$.
The basic concept behind this article is the use of Linear Approximation to calculate the value of the given linear function to an approximately accurate value.
Linear Approximation is a mathematical process in which the value of a given function is approximated or estimated at a certain point in the form of a line expression consisting of one real variable. The Linear Approximation is expressed by $L(x)$.
For a given function $f(x)$ consisting of one real variable, if it is differentiated, then as per Taylor’s theorem:
\[f\left(x\right)\ =\ f\left(a\right)\ +\ f^\prime\left(a\right)\left(x-a\right)\ +\ R\]
In this expression, $R$ is the Remainder Term which is not considered during the Linear Approximation of a function. So for a given function $f(x)$ consisting of one real variable, the Linear Approximation will be:
\[L\left(x\right)\ \approx\ f\left(a\right)\ +\ f^\prime\left(a\right)\left(x\ -\ a\right)\]
Expert Answer
Given function is:
\[f(x)=\sqrt{4-x}\]
And:
\[a=0\]
In order to find the Linear Approximation $L(x)$, we need to find the value for $f(a)$ and $f^\prime(x)$ as follows:
\[f(x)=\sqrt{4-x}\]
So $f(a)$ at $x=a$ will be:
\[f(a)=\sqrt{4-a}\]
\[f(0)=\sqrt{4-0}\]
\[f(0)=\sqrt4\]
\[f(0)=2\]
$f^\prime(x)$ will be calculated as follows:
\[f^\prime(x)=\frac{d}{dx}\sqrt{4-x}\]
\[f^\prime(x)=-\frac{1}{2\sqrt{4-x}}\]
So $f^\prime(x)$ at $x=a$ will be:
\[f^\prime(a)=-\frac{1}{2\sqrt{4-a}}\]
\[f^\prime(0)=-\frac{1}{2\sqrt{4-0}}\]
\[f^\prime(0)=-\frac{1}{2\sqrt4}\]
\[f^\prime(0)=-\frac{1}{2\times2}=-\frac{1}{4}\]
As we know that the expression for Linear Approximation $L(x)$ is given as follows:
\[L\left(x\right)\ \approx\ f\left(a\right)\ +\ f^\prime\left(a\right)\left(x\ -\ a\right)\]
Substituting the values for $f(a)$ and $f^\prime(x)$ in above equation at $a=0$:
\[L\left(x\right)\ \approx\ f\left(0\right)\ +\ f^\prime\left(0\right)\left(x\ -\ 0\right)\]
\[L\left(x\right)\ \approx\ 2\ +\ (-\frac{1}{4})\left(x\right)\]
\[L\left(x\right)\ \approx\ 2\ -\ \frac{1}{4}x\]
For the given function $f(x)=\sqrt{4-x}$ will be equal to $\sqrt{3.9}$ as follows:
\[\sqrt{4-x}=\sqrt{3.9}\]
\[4-x=3.9\]
\[x=0.1\]
Hence, Linear Approximation for $\sqrt{3.9}$ at $x=0.1$ is as follows:
\[L\left(x\right)\ \approx\ 2\ -\ \frac{1}{4}x\]
\[L\left(0.1\right)\ \approx\ 2\ -\ \frac{1}{4}(0.1)\]
\[L\left(0.1\right)\ \approx\ 1.9750\]
For the given function $f(x)=\sqrt{4-x}$ will be equal to $\sqrt{3.99}$ as follows:
\[\sqrt{4-x}=\sqrt{3.99}\]
\[4-x=3.99\]
\[x=0.01\]
Hence, Linear Approximation for $\sqrt{3.99}$ at $x=0.01$ is as follows:
\[L\left(x\right)\ \approx\ 2\ -\ \frac{1}{4}x\]
\[L\left(0.1\right)\ \approx\ 2\ -\ \frac{1}{4}(0.01)\]
\[L\left(0.1\right)\ \approx\ 1.9975\]
Numerical Result
The Linear Approximation for the linear function $f(x)=\sqrt{4-x}$ at $a=0$ is:
\[L\left(x\right)\ \approx\ 2\ -\ \frac{1}{4}x\]
The Linear Approximation for $\sqrt{3.9}$ at $x=0.1$ is as follows:
\[L\left(0.1\right)\ \approx\ 1.9750\]
The Linear Approximation for $\sqrt{3.99}$ at $=0.01$ is as follows:
\[L\left(0.1\right)\ \approx\ 1.9975\]
Example
For the given linear function as $f(x)=\sqrt x$, calculate the Linear Approximation at $a=9$.
Solution
Given function is:
\[f(x)=\sqrt x\]
And:
\[a=9\]
In order to find theLinear Approximation $L(x)$, we need to find the value for $f(a)$ and f^\prime(x) as follows:
\[f(x)=\sqrt x\]
So $f(a)$ at $x=a$ will be:
\[f(a)=\sqrt a\]
\[f(9)=\sqrt9\]
\[f(9)=3\]
$f^\prime(x)$ will be calculated as follows:
\[f^\prime(x)=\frac{d}{dx}\sqrt x\]
\[f^\prime(x)=\frac{1}{2\sqrt x}\]
So $f^\prime(x)$ at $x=a$ will be:
\[f^\prime(a)=\frac{1}{2\sqrt a}\]
\[f^\prime(9)=\frac{1}{2\sqrt 9}\]
\[f^\prime(9)=\frac{1}{2\times3}\]
\[f^\prime(9)=\frac{1}{6}\]
As we know, the expression for Linear Approximation $L(x)$ is given as follows:
\[L\left(x\right)\ \approx\ f\left(a\right)\ +\ f^\prime\left(a\right)\left(x\ -\ a\right)\]
Substituting the values for $f(a)$ and $f^\prime(x)$ in above equation at $a=9$:
\[L\left(x\right)\ \approx\ f\left(9\right)\ +\ f^\prime\left(9\right)\left(x\ -\ 9\right)\]
\[L\left(x\right)\ \approx\ 3\ +\ \frac{1}{6}\left(x-9\right)\]