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Question: How do you estimate the quantity using the linear approximation of \({{\left( 3.9 \right)}^{\dfrac{1...

How do you estimate the quantity using the linear approximation of (3.9)12{{\left( 3.9 \right)}^{\dfrac{1}{2}}}?

Explanation

Solution

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(x)=f(a)+f′(a)(x−a)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 (3.9)12{{\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(x)=f(a)+f′(a)(x−a)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(3.9)=3.9f\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 aa.
Let us consider a=4a=4 so we will get
⇒f(a)=f(4) ⇒f(4)=4 ⇒f(4)=2 \begin{aligned} & \Rightarrow f\left( a \right)=f\left( 4 \right) \\\ & \Rightarrow f\left( 4 \right)=\sqrt{4} \\\ & \Rightarrow f\left( 4 \right)=2 \\\ \end{aligned}
Now, we know that if f(x)=xf\left( x \right)=\sqrt{x} then f′(x)=12xf'\left( x \right)=\dfrac{1}{2\sqrt{x}}
So we get f′(4)=124f'\left( 4 \right)=\dfrac{1}{2\sqrt{4}}
Now, substituting the values in the formula of linear approximation we will get
⇒L(3.9)=2+124(3.9−4)\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
⇒L(3.9)=2+12×2(−0.1) ⇒L(3.9)=2+14(−0.1) ⇒L(3.9)=2+0.25(−0.1) ⇒L(3.9)=2−0.025 ⇒L(3.9)=1.975 \begin{aligned} & \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{aligned}
Hence we get (3.9)12≈1.975{{\left( 3.9 \right)}^{\dfrac{1}{2}}}\approx 1.975

Note: The point to be noted is that always choose the value of aa 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.