Given that x1 = 7.5 is an approximate root of e^x − 6x³ = 0, use the Newton–Raphson technique to find an improved value.

Question

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

Accepted Answer

[latex]\begin{aligned}x_1 & =7.5 & & \f(x) & =\mathrm{e}^x-6 x^3 & & f\left(x_1 ight)=-723 \f^{\prime}(x) & =\mathrm{e}^x-18 x^2 & & f^{\prime}\left(x_1 ight)=796\end{aligned}[/latex]

Using the Newton–Raphson technique the value of [latex]x_2[/latex] is found:

[latex]x_2=x_1-\frac{f\left(x_1 ight)}{f^{\prime}\left(x_1 ight)}=7.5-\frac{(-723)}{796}=8.41[/latex]

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

Chapter 12

Q. 12.6

Q. 12.6

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