Question

write the formula for newtons method and use the given initial approximation to compute the approximations ( x_1 ) and ( x_2 ).
( f(x)=2 e^{-x}-7 x, x_0=ln (4) )
( 2 e^{-x_n}-7 x_n )
( \bigcirc ) b. ( x_{n+1}=x_n+\frac{2 e^{-x_n}-7 x_n}{-2 e^{-x_n}-7} )
( \bigcirc ) c. ( x_{n+1}=x_n+\frac{2 e^{-x_n}+7 x_n}{2 e^{-x_n}+7} )
( \bigcirc ) d. ( x_{n+1}=x_n-\frac{2 e^{-x_n}-7 x_n}{-2 e^{-x_n}-7} )
use the given initial approximation to compute the approximations ( x_1 ) and ( x_2 ).
( x_1=square )
(do not round until the final answer. then round to six decimal places as needed)

Explanation:

Step1: Recall Newton's method formula

Newton's method formula is \(x_{n + 1}=x_{n}-\frac{f(x_{n})}{f^{\prime}(x_{n})}\). Given \(f(x)=2e^{-x}-7x\), find its derivative \(f^{\prime}(x)\).
Using the rules of differentiation (\((e^{ax})^\prime = ae^{ax}\) and \((bx)^\prime=b\)), we have \(f^{\prime}(x)=- 2e^{-x}-7\).
So \(x_{n + 1}=x_{n}-\frac{2e^{-x_{n}}-7x_{n}}{-2e^{-x_{n}}-7}\), which is option D.

Step2: Compute \(x_1\)

Given \(x_0 = \ln(4)\approx1.386294\).
First, calculate \(f(x_0)=2e^{-\ln(4)}-7\ln(4)\).
Since \(e^{-\ln(4)}=\frac{1}{e^{\ln(4)}}=\frac{1}{4}\), then \(f(x_0)=2\times\frac{1}{4}-7\ln(4)=\frac{1}{2}-7\ln(4)\approx0.5 - 7\times1.386294=0.5-9.704058=-9.204058\).
Next, calculate \(f^{\prime}(x_0)=-2e^{-\ln(4)}-7=-2\times\frac{1}{4}-7=-\frac{1}{2}-7=-7.5\).
Then \(x_1=x_0-\frac{f(x_0)}{f^{\prime}(x_0)}=\ln(4)-\frac{-9.204058}{-7.5}\approx1.386294 - 1.227208 = 0.159086\).

Step3: Compute \(x_2\)

Now \(x_1 = 0.159086\).
Calculate \(f(x_1)=2e^{-0.159086}-7\times0.159086\).
\(e^{-0.159086}\approx0.8537\), so \(f(x_1)=2\times0.8537-7\times0.159086 = 1.7074-1.113602=0.593798\).
Calculate \(f^{\prime}(x_1)=-2e^{-0.159086}-7\approx-2\times0.8537-7=-1.7074 - 7=-8.7074\).
Then \(x_2=x_1-\frac{f(x_1)}{f^{\prime}(x_1)}=0.159086-\frac{0.593798}{-8.7074}\approx0.159086 + 0.068193=0.227279\).

Answer:

\(x_1 = 0.159086\)