Question

a. write the equation of the line that represents the linear approximation to the following function at the given point a
b. use the linear approximation to estimate the given quantity.
c. compute the percent error in the approximation, \\( \frac { 100 | \text { approximation } - \text { exact } | } { | \text { exact } | } \\), where the exact value is given by a calculator.
\\( f ( x ) = \ln ( 1 + x ), a = 0, f ( 0.6 ) \\)
a. \\( l ( x ) = x \\)
b. using the linear approximation, \\( f ( 0.6 ) \approx 0.6 \\)
(round to one decimal place as needed )
c. the percent error in the approximation is \\( \square \\% \\)
(round to two decimal places as needed )

Explanation:

Step1: Find the exact value of \( f(0.6) \)

The function is \( f(x)=\ln(1 + x) \). When \( x = 0.6 \), the exact value \( f(0.6)=\ln(1 + 0.6)=\ln(1.6)\approx0.4700036292 \)

Step2: Calculate the percent - error formula

The approximation is \( L(0.6)=0.6 \). The percent - error formula is \( E=\frac{100| \text{approximation}-\text{exact}|}{|\text{exact}|}\)

Substitute the values: \( \text{approximation}=0.6 \), \( \text{exact}\approx0.4700036292 \)

\( E=\frac{100|0.6 - 0.4700036292|}{|0.4700036292|}\)

First, calculate \(|0.6 - 0.4700036292|=0.1299963708\)

Then \( E=\frac{100\times0.1299963708}{0.4700036292}\)

\( E=\frac{12.99963708}{0.4700036292}\approx27.66\)

Answer:

\(27.66\)