Compare Newton-Raphson with linear interpolation

A question is this type if and only if it asks to find approximations using both Newton-Raphson and linear interpolation methods on the same problem.

3 questions

Edexcel F1 2018 January Q1
1. $$f ( x ) = 3 x ^ { 2 } - \frac { 5 } { 3 \sqrt { x } } - 6 , \quad x > 0$$ The single root \(\alpha\) of the equation \(\mathrm { f } ( x ) = 0\) lies in the interval [1.5, 1.6].
  1. Taking 1.5 as a first approximation to \(\alpha\), apply the Newton-Raphson process once to \(\mathrm { f } ( x )\) to obtain a second approximation to \(\alpha\). Give your answer to 3 decimal places.
    [0pt]
  2. Use linear interpolation once on the interval [1.5, 1.6] to find another approximation to \(\alpha\). Give your answer to 3 decimal places.
Edexcel F1 2014 June Q4
4. $$\mathrm { f } ( x ) = x ^ { \frac { 3 } { 2 } } - 3 x ^ { \frac { 1 } { 2 } } - 3 , \quad x > 0$$ Given that \(\alpha\) is the only real root of the equation \(\mathrm { f } ( x ) = 0\),
  1. show that \(4 < \alpha < 5\)
  2. Taking 4.5 as a first approximation to \(\alpha\), apply the Newton-Raphson procedure once to \(\mathrm { f } ( x )\) to find a second approximation to \(\alpha\), giving your answer to 3 decimal places.
    [0pt]
  3. Use linear interpolation once on the interval [4,5] to find another approximation to \(\alpha\), giving your answer to 3 decimal places.
Edexcel FP1 2011 January Q3
3. $$f ( x ) = 5 x ^ { 2 } - 4 x ^ { \frac { 3 } { 2 } } - 6 , \quad x \geqslant 0$$ The root \(\alpha\) of the equation \(\mathrm { f } ( x ) = 0\) lies in the interval \([ 1.6,1.8 ]\).
  1. Use linear interpolation once on the interval \([ 1.6,1.8 ]\) to find an approximation to \(\alpha\). Give your answer to 3 decimal places.
  2. Differentiate \(\mathrm { f } ( x )\) to find \(\mathrm { f } ^ { \prime } ( x )\).
  3. Taking 1.7 as a first approximation to \(\alpha\), apply the Newton-Raphson process once to \(\mathrm { f } ( x )\) to obtain a second approximation to \(\alpha\). Give your answer to 3 decimal places.