Newton-Raphson with Other Methods

1. $$\mathrm { f } ( x ) = 6 \sqrt { x } - x ^ { 2 } - \frac { 1 } { 2 x } , \quad x > 0$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) in the interval \([ 3,4 ]\).
2. Taking 3 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.
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3. Use linear interpolation once on the interval [3,4] to find another approximation to \(\alpha\). Give your answer to 3 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.

3. $$\mathrm { f } ( x ) = x ^ { 3 } - 5 \sqrt { x } - 4 x + 7 \quad x \geqslant 0$$ The equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) in the interval \([ 0.25,1 ]\)

1. Use linear interpolation once on the interval [ \(0.25,1\) ] to determine an approximation to \(\alpha\), giving your answer to 3 decimal places. The equation \(\mathrm { f } ( x ) = 0\) has another root \(\beta\) in the interval [1.5, 2.5]
2. Determine \(\mathrm { f } ^ { \prime } ( x )\)
3. Hence, using \(x _ { 0 } = 1.75\) as a first approximation to \(\beta\), apply the Newton-Raphson process once to \(\mathrm { f } ( x )\) to determine a second approximation to \(\beta\), 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.

   VIIN SIHILNI III IM ION OC

   VIAV SIHI NI III HM ION OO

   VIAV SIHI NI III IM I ON OC

2. $$f ( x ) = 5 x ^ { 2 } - 4 x ^ { \frac { 3 } { 2 } } - 6 , \quad x \geqslant 0$$ 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. 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.

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.

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.

2 A curve satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 ^ { x }$$ Starting at the point \(( 1,4 )\) on the curve, use a step-by-step method with a step length of 0.01 to estimate the value of \(y\) at \(x = 1.02\). Give your answer to six significant figures.