Taking \(1.45\) as a first approximation to \(\alpha\), apply the Newton-Raphson procedure once to \(\mathrm{f}(x)=x^{3}-\frac{7}{x}+2\) to obtain a second approximation to \(\alpha\), giving your answer to 3 decimal places.

Question

\[
\mathrm{f}(x)=x^{3}-\frac{7}{x}+2, \quad x>0
\]
Taking \(1.45\) as a first approximation to \(\alpha\), apply the Newton-Raphson procedure once to \(\mathrm{f}(x)=x^{3}-\frac{7}{x}+2\) to obtain a second approximation to \(\alpha\), giving your answer to 3 decimal places.

Answer 1

\begin{aligned}
&\mathrm{f}(1.4)=\ldots \text { and } \mathrm{f}(1.5)=\ldots \\
&\mathrm{f}(1.4)=-0.256 \text { (or }-\frac{32}{125} \text { ), } \mathrm{f}(1.5)=0.708 \ldots \text { (or } \frac{17}{24} \text { ) } \\
&\text { Change of sign, } \therefore \text { root } \\
&\text { Alternative method: } \\
&\text { Graphical method could earn M1 if } 1.4 \text { and } 1.5 \text { are both indicated } \\
&\text { A1 then needs correct graph and conclusion, i.e. change of sign } \therefore \text { root } \\
&\text { Note } \\
&\text { M1: Some attempt at two evaluations } \\
&\text { A1: needs accuracy to } 1 \text { figure truncated or rounded and conclusion } \\
&\text { including sign change indicated (One figure accuracy sufficient) } \\
&\mathrm{f}(1.45)=0.221 \ldots \quad \text { or } 0.2 \quad[\therefore \text { root is in }[1.4,1.45]] \\
&\mathrm{f}(1.425)=-0.018 \ldots \text { or }-0.019 \text { or }-0.02 \\
&\therefore \text { root is in }[1.425,1.45] \\
&\text { Note } \\
&\text { M1: See } \mathrm{f}(1.45) \text { attempted and positive } \\
&\text { M1: See } \mathrm{f}(1.425) \text { attempted and negative } \\
&\text { Al: is cso }-\text { any slips in numerical work are penalised here even } \\
&\text { if correct region found. } \\
&\text { Answer may be written as } 1.425 \leq \alpha \leq 1.45 \text { or } 1.425<\alpha<1.45 \text { or }
\end{aligned}

\((1.425,1.45)\) must be correct way round. Between is sufficient. There is no credit for linear interpolation.