
How do you estimate the quantity using the linear approximation of ${{\left( 3.9 \right)}^{\dfrac{1}{2}}}$?
Answer
473.7k+ views
Hint: We will use the linear approximation formula to find the linear approximation of the given square root number. The linear approximation formula is given by $L\left( x \right)=f\left( a \right)+f'\left( a \right)\left( x-a \right)$ where a is the value we need to consider close to the given value.
Complete step-by-step solution:
We have been given a quantity ${{\left( 3.9 \right)}^{\dfrac{1}{2}}}$.
We have to estimate the given quantity using the linear approximation.
We know that the linear approximation of a function is given as $L\left( x \right)=f\left( a \right)+f'\left( a \right)\left( x-a \right)$.
Here in this question we have to estimate the function $f\left( 3.9 \right)=\sqrt{3.9}$
Now, first we need to choose a value closer to the given number which is the value of $a$.
Let us consider $a=4$ so we will get
$\begin{align}
& \Rightarrow f\left( a \right)=f\left( 4 \right) \\
& \Rightarrow f\left( 4 \right)=\sqrt{4} \\
& \Rightarrow f\left( 4 \right)=2 \\
\end{align}$
Now, we know that if $f\left( x \right)=\sqrt{x}$ then $f'\left( x \right)=\dfrac{1}{2\sqrt{x}}$
So we get $f'\left( 4 \right)=\dfrac{1}{2\sqrt{4}}$
Now, substituting the values in the formula of linear approximation we will get
$\Rightarrow L\left( 3.9 \right)=2+\dfrac{1}{2\sqrt{4}}\left( 3.9-4 \right)$
Simplifying the above obtained equation we will get
$\begin{align}
& \Rightarrow L\left( 3.9 \right)=2+\dfrac{1}{2\times 2}\left( -0.1 \right) \\
& \Rightarrow L\left( 3.9 \right)=2+\dfrac{1}{4}\left( -0.1 \right) \\
& \Rightarrow L\left( 3.9 \right)=2+0.25\left( -0.1 \right) \\
& \Rightarrow L\left( 3.9 \right)=2-0.025 \\
& \Rightarrow L\left( 3.9 \right)=1.975 \\
\end{align}$
Hence we get ${{\left( 3.9 \right)}^{\dfrac{1}{2}}}\approx 1.975$
Note: The point to be noted is that always choose the value of $a$ closer to the given number such as the square root of that number can be easily calculated. The method of approximation is useful to find the value of a function at any point. We can easily find the value of a function at any particular point.
Complete step-by-step solution:
We have been given a quantity ${{\left( 3.9 \right)}^{\dfrac{1}{2}}}$.
We have to estimate the given quantity using the linear approximation.
We know that the linear approximation of a function is given as $L\left( x \right)=f\left( a \right)+f'\left( a \right)\left( x-a \right)$.
Here in this question we have to estimate the function $f\left( 3.9 \right)=\sqrt{3.9}$
Now, first we need to choose a value closer to the given number which is the value of $a$.
Let us consider $a=4$ so we will get
$\begin{align}
& \Rightarrow f\left( a \right)=f\left( 4 \right) \\
& \Rightarrow f\left( 4 \right)=\sqrt{4} \\
& \Rightarrow f\left( 4 \right)=2 \\
\end{align}$
Now, we know that if $f\left( x \right)=\sqrt{x}$ then $f'\left( x \right)=\dfrac{1}{2\sqrt{x}}$
So we get $f'\left( 4 \right)=\dfrac{1}{2\sqrt{4}}$
Now, substituting the values in the formula of linear approximation we will get
$\Rightarrow L\left( 3.9 \right)=2+\dfrac{1}{2\sqrt{4}}\left( 3.9-4 \right)$
Simplifying the above obtained equation we will get
$\begin{align}
& \Rightarrow L\left( 3.9 \right)=2+\dfrac{1}{2\times 2}\left( -0.1 \right) \\
& \Rightarrow L\left( 3.9 \right)=2+\dfrac{1}{4}\left( -0.1 \right) \\
& \Rightarrow L\left( 3.9 \right)=2+0.25\left( -0.1 \right) \\
& \Rightarrow L\left( 3.9 \right)=2-0.025 \\
& \Rightarrow L\left( 3.9 \right)=1.975 \\
\end{align}$
Hence we get ${{\left( 3.9 \right)}^{\dfrac{1}{2}}}\approx 1.975$
Note: The point to be noted is that always choose the value of $a$ closer to the given number such as the square root of that number can be easily calculated. The method of approximation is useful to find the value of a function at any point. We can easily find the value of a function at any particular point.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Trending doubts
Explain zero factorial class 11 maths CBSE

What is the type of food and mode of feeding of the class 11 biology CBSE

State the laws of reflection of light

Earth rotates in which direction A East to west B West class 11 physics CBSE

What organs are located on the left side of your body class 11 biology CBSE

Draw the molecular orbital diagram of N2N2 + N2 Write class 11 chemistry CBSE
