Question 12.8: Given that x1 =7.5 is an approximate root of e^x - 6x³ = 0, ......
Given that x_1 =7.5 is an approximate root of e^x – 6x³ = 0, use the Newton-Raphson technique to find an improved value.
Step-by-Step
\begin{array}{rlrl} x_1 & =7.5 & \\ f(x) & =\mathrm{e}^x-6 x^3 & & f\left(x_1\right)=-723 \\ f^{\prime}(x) & =\mathrm{e}^x-18 x^2 & & f^{\prime}\left(x_1\right)=796 \end{array}
Using the Newton-Raphson technique the value of x_2 is found:
x_2=x_1-\frac{f\left(x_1\right)}{f^{\prime}\left(x_1\right)}=7.5-\frac{(-723)}{796}=8.41An improved estimate of the root of \mathrm{e}^x-6 x^3=0 \text { is } x=8.41 \text {. } To two decimal places the true answer is a = 8.05.
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