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Use L(x) to approximate the numbers √(3.9) and √(3.99). (Round your answers to four decimal places.)

– 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 $a=9$:

\[L\left(x\right)\ \approx\ f\left(9\right)\ +\ f^\prime\left(9\right)\left(x\ -\ 9\right)\]

\[L\left(x\right)\ \approx\ 3\ +\ \frac{1}{6}\left(x-9\right)\]

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