Question

write the formula for newtons method and use the given initial approximation to compute the approximations ( x_1 ) and ( x_2 ).
( f(x)=6 e^{-x}-23 x, x_0=ln (4) )
( \bigcirc mathrm{b} ). ( x_{n+1}=x_n+\frac{6 e^{-x_n}+23 x_n}{6 e^{-x_n}+23} )
( \bigcirc mathrm{c} ). ( x_{n+1}=x_n+\frac{6 e^{-x_n}-23 x_n}{-6 e^{-x_n}-23} )
( \bigcirc mathrm{d} ). ( x_{n+1}=x_n-\frac{6 e^{-x_n}-23 x_n}{-6 e^{-x_n}-23} )
use the given initial approximation to compute the approximations ( x_1 ) and ( x_2 ).
( x_1=0.1461 )
(do not round until the final answer. then round to six decimal places as needed.)
( x_2=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 \(f(x)=6e^{-x}-23x\), then \(f^{\prime}(x)=-6e^{-x}-23\). So \(x_{n + 1}=x_{n}-\frac{6e^{-x_{n}}-23x_{n}}{-6e^{-x_{n}}-23}\), which is option D.

Step2: Calculate \(x_1\)

We are given \(x_0 = \ln(4)\approx1.386294\). First, find \(f(x_0)=6e^{-\ln(4)}-23\ln(4)=6\times\frac{1}{4}-23\times1.386294 = 1.5-31.884762=-30.384762\). Then \(f^{\prime}(x_0)=-6e^{-\ln(4)}-23=-6\times\frac{1}{4}-23=-1.5 - 23=-24.5\). Using \(x_{1}=x_{0}-\frac{f(x_{0})}{f^{\prime}(x_{0})}\), \(x_{1}=1.386294-\frac{- 30.384762}{-24.5}=1.386294 - 1.2402=\approx0.146094\) (matches the given \(x_1\)).

Step3: Calculate \(x_2\)

Now, for \(n = 1\), \(x_1\approx0.146094\). Find \(f(x_1)=6e^{-0.146094}-23\times0.146094\). \(e^{-0.146094}\approx0.864\), so \(f(x_1)=6\times0.864-3.360162 = 5.184-3.360162 = 1.823838\). \(f^{\prime}(x_1)=-6e^{-0.146094}-23\approx-6\times0.864-23=-5.184-23=-28.184\). Then \(x_{2}=x_{1}-\frac{f(x_{1})}{f^{\prime}(x_{1})}=0.146094-\frac{1.823838}{-28.184}=0.146094 + 0.06471\approx0.210804\)

Answer:

\(x_{2}\approx0.210804\)