Q. 12.6

Given that   $x_1$  = 7.5 is an approximate root of   $e^x-6 x^3=0$,   use the Newton–Raphson technique to find an improved value.

Verified Solution

\begin{aligned}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{aligned}

Using the Newton–Raphson technique the value of  $x_2$  is found:

$x_2=x_1-\frac{f\left(x_1\right)}{f^{\prime}\left(x_1\right)}=7.5-\frac{(-723)}{796}=8.41$

An improved estimate of the root of   $e^x-6 x^3=0$  is x = 8.41. To two decimal places the true answer is x = 8.05.