3.
$$\mathrm { f } ( x ) = x ^ { 3 } - \frac { 7 } { x } + 2 , \quad x > 0$$
- Show that \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) between 1.4 and 1.5
[0pt] - Starting with the interval [1.4,1.5], use interval bisection twice to find an interval of width 0.025 that contains \(\alpha\).
- 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.