Interval Bisection with Other Methods

1. In the interval \(2 < x < 3\), the equation

$$6 - x ^ { 2 } \cos \left( \frac { x } { 5 } \right) = 0 , \text { where } x \text { is measured in radians }$$ has exactly one root \(\alpha\).
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1. Starting with the interval [2,3], use interval bisection twice to find an interval of width 0.25 which contains \(\alpha\).
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2. Use linear interpolation once on the interval [2,3] to find an approximation to \(\alpha\). Give your answer to 2 decimal places.

5. $$f ( x ) = 30 + \frac { 7 } { \sqrt { x } } - x ^ { 5 } , \quad x > 0$$ The only real root, \(\alpha\), of the equation \(\mathrm { f } ( x ) = 0\) lies in the interval [2,2.1].
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1. Starting with the interval [2,2.1], use interval bisection twice to find an interval of width 0.025 that contains \(\alpha\).
2. Taking 2 as a first approximation to \(\alpha\), apply the Newton-Raphson process once to \(\mathrm { f } ( x )\) to find a second approximation to \(\alpha\), giving your answer to 2 decimal places.

7. $$f ( x ) = x ^ { \frac { 3 } { 2 } } + x - 3$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root, \(\alpha\), in the interval \([ 1,2 ]\)
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2. Starting with the interval [1, 2], use interval bisection twice to show that \(\alpha\) lies in the interval [1.25, 1.5]
3. 1. Determine \(\mathrm { f } ^ { \prime } ( x )\)
   2. Using 1.375 as a first approximation for \(\alpha\), apply the Newton-Raphson process once to \(\mathrm { f } ( x )\) to determine a second approximation for \(\alpha\), giving your answer to 3 decimal places.
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4. Use linear interpolation once on the interval [1.25,1.5] to obtain a different approximation for \(\alpha\), giving your answer to 3 decimal places.

2. $$f ( x ) = 3 x ^ { 2 } - \frac { 11 } { x ^ { 2 } }$$

1. Write down, to 3 decimal places, the value of \(\mathrm { f } ( 1.3 )\) and the value of \(\mathrm { f } ( 1.4 )\). The equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) between 1.3 and 1.4
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2. Starting with the interval [1.3, 1.4], use interval bisection to find an interval of width 0.025 which contains \(\alpha\).
3. Taking 1.4 as a first approximation to \(\alpha\), apply the Newton-Raphson procedure once to \(\mathrm { f } ( x )\) to obtain a second approximation to \(\alpha\), giving your answer to 3 decimal places.

2. (a) Show that \(\mathrm { f } ( x ) = x ^ { 4 } + x - 1\) has a real root \(\alpha\) in the interval [0.5, 1.0].
[0pt] (b) Starting with the interval [0.5, 1.0], use interval bisection twice to find an interval of width 0.125 which contains \(\alpha\).
(c) Taking 0.75 as a first approximation, apply the Newton Raphson process twice to \(\mathrm { f } ( x )\) to obtain an approximate value of \(\alpha\). Give your answer to 3 decimal places.

3. $$\mathrm { f } ( x ) = x ^ { 3 } - \frac { 7 } { x } + 2 , \quad x > 0$$

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

8. $$f ( x ) = x ^ { 3 } - 2 x - 3$$

1. Show that \(\mathrm { f } ( x ) = 0\) has a root, \(\alpha\), in the interval \([ 1,2 ]\).
2. Starting with the interval \([ 1,2 ]\), use interval bisection twice to find an interval of width 0.25 which contains \(\alpha\).
3. Using \(x _ { 0 } = 1.8\) 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 significant figures.

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

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) in the interval [2,3].
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2. Use linear interpolation once on the interval [2,3] to find an approximation to \(\alpha\). Give your answer to 3 decimal places.
3. The equation \(\mathrm { f } ( x ) = 0\) has another root \(\beta\) in the interval \([ - 1,0 ]\). Starting with this interval, use interval bisection to find an interval of width 0.25 which contains \(\beta\).

2. In the interval \(13 < x < 14\), the equation $$3 + x \sin \left( \frac { x } { 4 } \right) = 0 , \text { where } x \text { is measured in radians, }$$ has exactly one root, \(\alpha\).
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1. Starting with the interval [13,14], use interval bisection twice to find an interval of width 0.25 which contains \(\alpha\).
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2. Use linear interpolation once on the interval [13,14] to find an approximate value for \(\alpha\). Give your answer to 3 decimal places.

7 The equation $$24 x ^ { 3 } + 36 x ^ { 2 } + 18 x - 5 = 0$$ has one real root, \(\alpha\).

1. Show that \(\alpha\) lies in the interval \(0.1 < x < 0.2\).
2. Starting from the interval \(0.1 < x < 0.2\), use interval bisection twice to obtain an interval of width 0.025 within which \(\alpha\) must lie.
3. Taking \(x _ { 1 } = 0.2\) as a first approximation to \(\alpha\), use the Newton-Raphson method to find a second approximation, \(x _ { 2 }\), to \(\alpha\). Give your answer to four decimal places.
   (4 marks)