1.09a Sign change methods: locate roots

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.
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3. Use linear interpolation once on the interval [4,5] to find another approximation to \(\alpha\), giving your answer to 3 decimal places.

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.

6. $$f ( x ) = \frac { 2 \left( x ^ { 3 } + 3 \right) } { \sqrt { x } } - 9 , \quad x > 0$$ The equation \(\mathrm { f } ( x ) = 0\) has two real roots \(\alpha\) and \(\beta\), where \(0.4 < \alpha < 0.5\) and \(1.2 < \beta < 1.3\)

1. Taking 0.45 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.
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2. Use linear interpolation once on the interval [1.2, 1.3] to find an approximation to \(\beta\), giving your answer to 3 decimal places.

1. $$f ( x ) = x ^ { 3 } - \frac { 10 \sqrt { x } - 4 x } { x ^ { 2 } } \quad x > 0$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) in the interval[1.4,1.5]
2. Determine \(\mathrm { f } ^ { \prime } ( x )\) .
3. Using \(x _ { 0 } = 1.4\) as a first approximation to \(\alpha\) ,apply the Newton-Raphson procedure once to \(\mathrm { f } ( x )\) to calculate a second approximation to \(\alpha\) ,giving your answer to 3 decimal places. \(f ( x ) = x ^ { 3 } - \frac { 10 \sqrt { x } - 4 x } { x ^ { 2 } } \quad x > 0\)
   1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) in the interval[1.4,1.5]
   2. Determine \(\mathrm { f } ^ { \prime } ( x )\) .

2. $$f ( x ) = 10 - 2 x - \frac { 1 } { 2 \sqrt { x } } - \frac { 1 } { x ^ { 3 } } \quad x > 0$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) in the interval [0.4, 0.5]
2. Determine \(\mathrm { f } ^ { \prime } ( x )\).
3. Using \(x _ { 0 } = 0.5\) 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. The equation \(\mathrm { f } ( x ) = 0\) has another root \(\beta\) in the interval [4.8, 4.9]
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4. Use linear interpolation once on the interval [4.8, 4.9] to find an approximation to \(\beta\), giving your answer to 3 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 \([ 1,2 ]\) [0pt]
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. $$f ( x ) = 7 \sqrt { x } - \frac { 1 } { 2 } x ^ { 3 } - \frac { 5 } { 3 x } \quad x > 0$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root, \(\alpha\), in the interval [2.8, 2.9]
2. 1. Find \(\mathrm { f } ^ { \prime } ( x )\).
   2. Hence, using \(x _ { 0 } = 2.8\) as a first approximation to \(\alpha\), apply the Newton-Raphson procedure once to \(\mathrm { f } ( x )\) to calculate a second approximation to \(\alpha\), giving your answer to 3 decimal places.
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3. Use linear interpolation once on the interval [2.8, 2.9] to find another approximation to \(\alpha\). Give your answer to 3 decimal places.

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3. $$\mathrm { f } ( x ) = x ^ { 2 } + \frac { 3 } { x } - 1 , \quad x < 0$$ The only real root, \(\alpha\), of the equation \(\mathrm { f } ( x ) = 0\) lies in the interval \([ - 2 , - 1 ]\).

1. Taking - 1.5 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 2 decimal places.
2. Show that your answer to part (a) gives \(\alpha\) correct to 2 decimal places. \includegraphics[max width=\textwidth, alt={}, center]{38217fcb-8f26-49ac-9bb1-61c2f304006e-06_2250_51_317_1980}

5. $$f ( x ) = 3 \sqrt { } x + \frac { 18 } { \sqrt { } x } - 20$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) in the interval [1.1,1.2].
2. Find \(\mathrm { f } ^ { \prime } ( x )\).
3. Using \(x _ { 0 } = 1.1\) 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. $$f ( x ) = 3 x ^ { 2 } - \frac { 11 } { x ^ { 2 } }$$

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.

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.

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.

4. Given that \(\alpha\) is the only real root of the equation $$x ^ { 3 } - x ^ { 2 } - 6 = 0$$

1. show that \(2.2 < \alpha < 2.3\)
2. Taking 2.2 as a first approximation to \(\alpha\), apply the Newton-Raphson procedure once to \(\mathrm { f } ( x ) = x ^ { 3 } - x ^ { 2 } - 6\) to obtain a second approximation to \(\alpha\), giving your answer to 3 decimal places.
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3. Use linear interpolation once on the interval [2.2, 2.3] to find another approximation to \(\alpha\), giving 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.

1. $$f ( x ) = 3 ^ { x } + 3 x - 7$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) between \(x = 1\) and \(x = 2\).
2. Starting with the interval \([ 1,2 ]\), use interval bisection twice to find an interval of width 0.25 which contains \(\alpha\).

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.

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

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

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. $$\mathrm { f } ( x ) = x ^ { 3 } - \frac { 5 } { 2 x ^ { \frac { 3 } { 2 } } } + 2 x - 3 , \quad x > 0$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) in the interval [1.1,1.5].
2. Find f'(x).
3. Using \(x _ { 0 } = 1.1\) 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.

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.

2. $$f ( x ) = \frac { 3 } { 2 } x ^ { 2 } + \frac { 4 } { 3 x } + 2 x - 5 , \quad x < 0$$ The equation \(\mathrm { f } ( x ) = 0\) has a single root \(\alpha\).

1. Show that \(\alpha\) lies in the interval \([ - 3 , - 2.5 ]\)
2. Taking - 3 as a first approximation to \(\alpha\), apply the Newton-Raphson procedure once to \(\mathrm { f } ( x )\) to obtain a second approximation to \(\alpha\). Give your answer to 3 decimal places.
3. Use linear interpolation once on the interval \([ - 3 , - 2.5 ]\) to find another approximation to \(\alpha\), giving your answer to 3 decimal places.

4. \(f ( x ) = x ^ { 3 } - 4 x ^ { 2 } + 5 x - 3\) The equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) in the interval ( 2,3 ).

1. Use linear interpolation on the end points of this interval to obtain an approximation for \(\alpha\).
2. Taking 2.5 as a first approximation to \(\alpha\), apply the Newton - Raphson procedure once to \(\mathrm { f } ( x )\) to obtain a second approximation to \(\alpha\). Give your answer to 2 decimal places.