Question

How do you find the linear approximation of ${{\left( 1.999 \right)}^{4}}$?

Answer 1

Hint: From the given number ${{\left( 1.999 \right)}^{4}}$, we can define a function $y={{x}^{4}}$. Now, we can write the number $1.999$ in the simplified form to be equal to $\left( 2-0.001 \right)$. So we can take $x=2$ and $\Delta x=-0.001$. Then, substituting these values in the relation $\Delta y=\dfrac{dy}{dx}\Delta x$ we can determine the change in the variable $y$. Finally, putting these values in the relation $\Delta y={{\left( x+\Delta x \right)}^{4}}-{{x}^{4}}$, we will get the required approximation of the given number.

Complete step by step answer:
Let us define a function as
$y=f\left( x \right)={{x}^{4}}.........(i)$
Now, we know that the differentiation of the function ${{x}^{n}}$ is $n{{x}^{n-1}}$, so on differentiating both sides of the above equation with respect to $x$, we get
$\dfrac{dy}{dx}=4{{x}^{3}}........(ii)$
According to the question, we have to find out the linear approximation of ${{\left( 1.999 \right)}^{4}}$. Now, the number $1.999$ can be written as $\left( 2-0.001 \right)$.
So we let $x=2$ and $\Delta x=-0.001$.
Now, the change in $y$ can be written as
$\Rightarrow \Delta y=f\left( x+\Delta x \right)-f\left( x \right)$
From (i) we can write the above equation as
$\Rightarrow \Delta y={{\left( x+\Delta x \right)}^{4}}-{{x}^{4}}$
Substituting $x=2$ and $\Delta x=-0.001$ in the above equation, we get
$\begin{align}
  & \Rightarrow \Delta y={{\left( 2-0.001 \right)}^{4}}-{{\left( 2 \right)}^{4}} \\
 & \Rightarrow \Delta y={{\left( 1.999 \right)}^{4}}-{{\left( 2 \right)}^{4}} \\
\end{align}$
We know that ${{2}^{4}}=2\times 2\times 2\times 2=16$. Substituting this above, we get
$\Rightarrow \Delta y={{\left( 1.999 \right)}^{4}}-16$
Adding $16$ both sides, we get
\[\Rightarrow {{\left( 1.999 \right)}^{4}}=\Delta y+16.........(iii)\]
Now, since $\Delta x=-0.001\approx 0$, we can say that $\Delta x\approx dx$. This implies that $\Delta y\approx dy$.
Therefore, $\Delta y$ can be given by
$\Rightarrow \Delta y=\dfrac{dy}{dx}\Delta x$
Putting the equation (ii) in the above equation, we get
$\Rightarrow \Delta y=4{{x}^{3}}\Delta x$
Substituting $x=2$ and $\Delta x=-0.001$ in the above equation, we get
$\Rightarrow \Delta y=4{{\left( 2 \right)}^{3}}\left( -0.001 \right)$
We know that ${{2}^{3}}=2\times 2\times 2=8$. Substituting this above, we get
$\Rightarrow \Delta y=4\times 8\times \left( -0.001 \right)$
On solving we get
$\Rightarrow \Delta y=-0.032$
Substituting this value in the equation (iii), we get
\[\Rightarrow {{\left( 1.999 \right)}^{4}}=-0.032+16\]
On solving, we finally get
\[\Rightarrow {{\left( 1.999 \right)}^{4}}=15.968\]

Hence, the linear approximation of ${{\left( 1.999 \right)}^{4}}$ is equal to $15.968$.

Note: The sign of $\Delta x$ is very important, so be careful regarding the sign. It is negative in this case, and is equal to $-0.001$. Do not take it equal to $0.001$. Usually, the values of $x$ and $\Delta x$ are chosen in such a way so that the value of $y=f\left( x \right)$ becomes a perfect integer.