Question

use linear approximation, i.e. the tangent line, to approximate \frac{1}{1.004} as follows: let f(x)=\frac{1}{x} and find the equation of the tangent line to f(x) at a
ice\ point near 1.004. then use this to approximate \frac{1}{1.004}

Explanation:

Step1: Find the derivative of \(f(x)\)

The function \(f(x)=\frac{1}{x}=x^{-1}\). Using the power rule \((x^{n})^\prime = nx^{n - 1}\), we have \(f^\prime(x)=-x^{-2}=-\frac{1}{x^{2}}\)

Step2: Choose a “nice” point \(a\)

A nice point near \(x = 1.004\) is \(a = 1\). Calculate \(f(1)\) and \(f^\prime(1)\)

• \(f(1)=\frac{1}{1}=1\)
• \(f^\prime(1)=-\frac{1}{1^{2}}=- 1\)

Step3: Write the equation of the tangent line

The equation of the tangent line \(L(x)\) using the formula \(L(x)=f(a)+f^\prime(a)(x - a)\) with \(a = 1\) is \(L(x)=f(1)+f^\prime(1)(x - 1)\)
Substitute \(f(1)=1\) and \(f^\prime(1)=-1\) into the formula: \(L(x)=1-(x - 1)=2 - x\)

Step4: Approximate \(f(1.004)\)

We use the tangent - line approximation \(f(x)\approx L(x)\) for \(x\) near \(a\). Let \(x = 1.004\), then \(f(1.004)=\frac{1}{1.004}\approx L(1.004)\)
Substitute \(x = 1.004\) into \(L(x)=2 - x\), we get \(L(1.004)=2-1.004 = 0.996\)

Answer:

\(0.996\)