Compare Newton-Raphson with linear interpolation

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.

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.
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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.

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

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) in the interval [2.5, 3.5]
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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.

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.

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.

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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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.

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.

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.

1.(i) $$f ( x ) = x ^ { 3 } + 4 x - 6$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) in the interval[1,1.5]
2. Taking 1.5 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. Show your working clearly.
   (ii) $$g ( x ) = 4 x ^ { 2 } + x - \tan x$$ where \(x\) is measured in radians. The equation \(\mathrm { g } ( x ) = 0\) has a single root \(\beta\) in the interval[1.4,1.5]
   Use linear interpolation on the values at the end points of this interval to obtain an approximation to \(\beta\) .Give your answer to 3 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}