Linear Interpolation Only

$$f ( x ) = x - 4 - \cos ( 5 \sqrt { x } ) \quad x > 0$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) in the interval [2.5, 3.5]
   [0pt]
2. Use linear interpolation once on the interval [2.5, 3.5] to find an approximation to \(\alpha\), giving your answer to 2 decimal places.
   (ii) $$\operatorname { g } ( x ) = \frac { 1 } { 10 } x ^ { 2 } - \frac { 1 } { 2 x ^ { 2 } } + x - 11 \quad x > 0$$
3. Determine \(\mathrm { g } ^ { \prime } ( x )\). The equation \(\mathrm { g } ( x ) = 0\) has a root \(\beta\) in the interval [6,7]
4. Using \(x _ { 0 } = 6\) as a first approximation to \(\beta\), apply the Newton-Raphson procedure once to \(\mathrm { g } ( x )\) to find a second approximation to \(\beta\), giving your answer to 3 decimal places.

6. $$f ( x ) = \tan \left( \frac { x } { 2 } \right) + 3 x - 6 , \quad - \pi < x < \pi$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) in the interval \([ 1,2 ]\).
2. Use linear interpolation once on the interval \([ 1,2 ]\) to find an approximation to \(\alpha\). Give your answer to 2 decimal places.

2. $$\mathrm { f } ( x ) = \cos \left( x ^ { 2 } \right) - x + 3 , \quad 0 < x < \pi$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) in the interval \([ 2.5,3 ]\).
   [0pt]
2. Use linear interpolation once on the interval [2.5,3] to find an approximation for \(\alpha\), giving your answer to 2 decimal places.

4. $$f ( x ) = 2 ^ { x } - 6 x$$ The equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) in the interval [4,5]. Using the end points of this interval find, by linear interpolation, an approximation to \(\alpha\).

1

1. Show that the equation $$x ^ { 3 } + 2 x - 2 = 0$$ has a root between 0.5 and 1 .
2. Use linear interpolation once to find an estimate of this root. Give your answer to two decimal places.

5 The diagram below (not to scale) shows a part of a curve \(y = \mathrm { f } ( x )\) which passes through the points \(A ( 2 , - 10 )\) and \(B ( 5,22 )\).

1. 1. On the diagram, draw a line which illustrates the method of linear interpolation for solving the equation \(\mathrm { f } ( x ) = 0\). The point of intersection of this line with the \(x\)-axis should be labelled \(P\).
   2. Calculate the \(x\)-coordinate of \(P\). Give your answer to one decimal place.
2. 1. On the same diagram, draw a line which illustrates the Newton-Raphson method for solving the equation \(\mathrm { f } ( x ) = 0\), with initial value \(x _ { 1 } = 2\). The point of intersection of this line with the \(x\)-axis should be labelled \(Q\).
   2. Given that the gradient of the curve at \(A\) is 8 , calculate the \(x\)-coordinate of \(Q\). Give your answer as an exact decimal.
      \includegraphics[max width=\textwidth, alt={}, center]{f9345653-d426-4350-bf1d-901506211078-3_876_1063_1779_523}

9. $$f ( x ) = 2 \sin 2 x + x - 2 .$$ The root \(\alpha\) of the equation \(\mathrm { f } ( x ) = 0\) lies in the interval \([ 2 , \pi ]\).
Using the end points of this interval find, by linear interpolation, an approximation to \(\alpha\).
[0pt] [*P4 January 2003 Qn 4]

17. $$f ( x ) = 2 ^ { x } + x - 4$$ The equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) in the interval [1,2].
Use linear interpolation on the values at the end points of this interval to find an approximation to \(\alpha\).
[0pt] [*P4 June 2004 Qn 2]