Find the local linear approximation of the function [tex]$f(x)=\sqrt{14+x}$[/tex] at [tex]$x_0=11$[/tex], and use it to approximate [tex]$\sqrt{24.9}$[/tex] and [tex]$\sqrt{25.1}$[/tex].
(a) [tex]$f(x)=\sqrt{14+x} \approx$[/tex] $\square$
(b) [tex]$\sqrt{24.9} \approx$[/tex] $\square$
(c) [tex]$\sqrt{25.1} \approx$[/tex] $\square$
For parts (b) and (c), you should enter your answer as a fraction. If you enter a decimal, make sure that it is correct to at least six decimal places.
Answer (2)
Find the derivative of f ( x ) = 14 + x , which is f ′ ( x ) = 2 14 + x 1 .
Evaluate f ( 11 ) = 5 and f ′ ( 11 ) = 10 1 .
Write the local linear approximation: f ( x ) ≈ 5 + 10 1 ( x − 11 ) .
Approximate 24.9 ≈ 100 499 and 25.1 ≈ 100 501 .
Explanation
Problem Analysis and Setup We are asked to find the local linear approximation of the function f ( x ) = 14 + x at x 0 = 11 , and then use this approximation to estimate the values of 24.9 and 25.1 . The local linear approximation is given by the formula:
f ( x ) ≈ f ( x 0 ) + f ′ ( x 0 ) ( x − x 0 )
Finding the Derivative and Evaluating at x_0 First, we need to find the derivative of f ( x ) . Using the power rule and chain rule, we have:
f ′ ( x ) = 2 14 + x 1
Next, we evaluate f ( x 0 ) and f ′ ( x 0 ) at x 0 = 11 :
f ( 11 ) = 14 + 11 = 25 = 5
f ′ ( 11 ) = 2 14 + 11 1 = 2 25 1 = 2 ⋅ 5 1 = 10 1
Writing the Local Linear Approximation Now we can write the local linear approximation:
f ( x ) ≈ f ( 11 ) + f ′ ( 11 ) ( x − 11 ) = 5 + 10 1 ( x − 11 )
Approximating sqrt(24.9) To approximate 24.9 , we need to find f ( 10.9 ) :
f ( 10.9 ) ≈ 5 + 10 1 ( 10.9 − 11 ) = 5 + 10 1 ( − 0.1 ) = 5 − 0.01 = 4.99
As a fraction, this is:
4.99 = 100 499
Approximating sqrt(25.1) To approximate 25.1 , we need to find f ( 11.1 ) :
f ( 11.1 ) ≈ 5 + 10 1 ( 11.1 − 11 ) = 5 + 10 1 ( 0.1 ) = 5 + 0.01 = 5.01
As a fraction, this is:
5.01 = 100 501
Final Answer Therefore, the local linear approximation of f ( x ) = 14 + x at x 0 = 11 is:
f ( x ) ≈ 5 + 10 1 ( x − 11 )
The approximation of 24.9 is 100 499 , and the approximation of 25.1 is 100 501 .
Examples
Linear approximation is incredibly useful in fields like physics and engineering. Imagine you're designing a bridge and need to calculate the expansion of a metal beam due to a slight temperature increase. Instead of using complex thermal expansion formulas for every tiny temperature change, you can use a linear approximation around the average temperature to quickly estimate the expansion. This simplifies calculations and provides a good enough estimate for many practical purposes, saving time and resources.
The local linear approximation of f ( x ) = 14 + x at x 0 = 11 is f ( x ) ≈ 5 + 10 1 ( x − 11 ) . Using this, 24.9 approximates to 100 499 and 25.1 approximates to 100 501 .
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