Question 17:
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(a)
Show that the equation \(x^3 + 2x - 6 = 0\) has a solution between \(x = 1\) and \(x = 2\).
(2 marks)
(b)
Show that the equation \(x^3 + 2x - 6 = 0\) can be rearranged to give \(x = \tfrac{6}{x^2 + 2}\).
(1 mark)
(c)
Starting with \(x_0 = 1.45\), use the iteration formula \(x_{n+1} = \tfrac{6}{x_n^2 + 2}\) twice to find an estimate for the solution of \(x^3 + 2x - 6 = 0\). Give your answer correct to 4 decimal places.
(3 marks)
(a) Answer:
\( \text{A root lies in }(1,2). \)
Explanation:
\[\begin{aligned}
&\text{Let } f(x)=x^3+2x-6.\\
&f(1)=1+2-6=-3<0,\quad f(2)=8+4-6=6>0.\\
&\text{Since }f\text{ changes sign on }[1,2],\text{ a root lies in }(1,2).
\end{aligned}\]
(b) Answer:
\( x = \dfrac{6}{x^2 + 2} \)
Explanation:
\[\begin{aligned}
&x^3+2x-6=0\\
&x^3+2x=6\\
&x(x^2+2)=6\\
&\therefore\; x=\dfrac{6}{x^2+2}.
\end{aligned}\]
(c) Answer:
\( 1.4496 \)
Explanation:
\[\begin{aligned}
&x_0=1.45\\
&x_1=\frac{6}{x_0^2+2}=\frac{6}{1.45^2+2}\approx 1.4625\\
&x_2=\frac{6}{x_1^2+2}=\frac{6}{1.4625^2+2}\approx 1.4496349\\
&\text{Solution }\approx 1.4496\;\text{(to 4 d.p.).}
\end{aligned}\]
