1.09a Sign change methods: locate roots

1. $$f ( x ) = 2 x ^ { 3 } + x - 10$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) in the interval \([ 1.5,2 ]\) The only real root of \(\mathrm { f } ( x ) = 0\) is \(\alpha\) The iterative formula $$x _ { n + 1 } = \left( 5 - \frac { 1 } { 2 } x _ { n } \right) ^ { \frac { 1 } { 3 } } , \quad x _ { 0 } = 1.5$$ can be used to find an approximate value for \(\alpha\)
2. Calculate \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\), giving your answers to 4 decimal places.
3. By choosing a suitable interval, show that \(\alpha = 1.6126\) correct to 4 decimal places.

1. $$f ( x ) = x ^ { 5 } + x ^ { 3 } - 12 x ^ { 2 } - 8 , \quad x \in \mathbb { R }$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) can be written as $$x = \sqrt [ 3 ] { \frac { 4 \left( 3 x ^ { 2 } + 2 \right) } { x ^ { 2 } + 1 } }$$
2. Use the iterative formula $$x _ { n + 1 } = \sqrt [ 3 ] { \frac { 4 \left( 3 x _ { n } ^ { 2 } + 2 \right) } { x _ { n } ^ { 2 } + 1 } }$$ with \(x _ { 0 } = 2\), to find \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\) giving your answers to 3 decimal places. The equation \(\mathrm { f } ( x ) = 0\) has a single root, \(\alpha\).
3. By choosing a suitable interval, prove that \(\alpha = 2.247\) to 3 decimal places.

5. $$f ( x ) = 2 x ^ { 3 } - x - 4$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) can be written as $$x = \sqrt { \left( \frac { 2 } { x } + \frac { 1 } { 2 } \right) }$$ The equation \(2 x ^ { 3 } - x - 4 = 0\) has a root between 1.35 and 1.4.
2. Use the iteration formula $$x _ { n + 1 } = \sqrt { } \left( \frac { 2 } { x _ { n } } + \frac { 1 } { 2 } \right)$$ with \(x _ { 0 } = 1.35\), to find, to 2 decimal places, the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\). The only real root of \(\mathrm { f } ( x ) = 0\) is \(\alpha\).
3. By choosing a suitable interval, prove that \(\alpha = 1.392\), to 3 decimal places.

7. $$f ( x ) = x ^ { 4 } - 4 x - 8$$

1. Show that there is a root of \(\mathrm { f } ( x ) = 0\) in the interval \([ - 2 , - 1 ]\).
2. Find the coordinates of the turning point on the graph of \(y = \mathrm { f } ( x )\).
3. Given that \(\mathrm { f } ( x ) = ( x - 2 ) \left( x ^ { 3 } + a x ^ { 2 } + b x + c \right)\), find the values of the constants, \(a , b\) and \(c\).
4. In the space provided on page 21, sketch the graph of \(y = \mathrm { f } ( x )\).
5. Hence sketch the graph of \(y = | \mathrm { f } ( x ) |\).

3. $$\mathrm { f } ( x ) = \ln ( x + 2 ) - x + 1 , \quad x > - 2 , x \in \mathbb { R } .$$

1. Show that there is a root of \(\mathrm { f } ( x ) = 0\) in the interval \(2 < x < 3\).
2. Use the iterative formula $$x _ { n + 1 } = \ln \left( x _ { n } + 2 \right) + 1 , x _ { 0 } = 2.5$$ to calculate the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\) giving your answers to 5 decimal places.
3. Show that \(x = 2.505\) is a root of \(\mathrm { f } ( x ) = 0\) correct to 3 decimal places.

5. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the curve with equation \(y = \mathrm { f } ( x )\), where $$\mathrm { f } ( x ) = ( 8 - x ) \ln x , \quad x > 0$$ The curve cuts the \(x\)-axis at the points \(A\) and \(B\) and has a maximum turning point at \(Q\), as shown in Figure 1.

1. Write down the coordinates of \(A\) and the coordinates of \(B\).
2. Find f'(x).
3. Show that the \(x\)-coordinate of \(Q\) lies between 3.5 and 3.6
4. Show that the \(x\)-coordinate of \(Q\) is the solution of $$x = \frac { 8 } { 1 + \ln x }$$ To find an approximation for the \(x\)-coordinate of \(Q\), the iteration formula $$x _ { n + 1 } = \frac { 8 } { 1 + \ln x _ { n } }$$ is used.
5. Taking \(x _ { 0 } = 3.55\), find the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\). Give your answers to 3 decimal places.

6. $$f ( x ) = x ^ { 2 } - 3 x + 2 \cos \left( \frac { 1 } { 2 } x \right) , \quad 0 \leqslant x \leqslant \pi$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a solution in the interval \(0.8 < x < 0.9\) The curve with equation \(y = \mathrm { f } ( x )\) has a minimum point \(P\).
2. Show that the \(x\)-coordinate of \(P\) is the solution of the equation $$x = \frac { 3 + \sin \left( \frac { 1 } { 2 } x \right) } { 2 }$$
3. Using the iteration formula $$x _ { n + 1 } = \frac { 3 + \sin \left( \frac { 1 } { 2 } x _ { n } \right) } { 2 } , \quad x _ { 0 } = 2$$ find the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\), giving your answers to 3 decimal places.
4. By choosing a suitable interval, show that the \(x\)-coordinate of \(P\) is 1.9078 correct to 4 decimal places.

2. $$\mathrm { g } ( x ) = \mathrm { e } ^ { x - 1 } + x - 6$$

1. Show that the equation \(\mathrm { g } ( x ) = 0\) can be written as $$x = \ln ( 6 - x ) + 1 , \quad x < 6$$ The root of \(\mathrm { g } ( x ) = 0\) is \(\alpha\).
   The iterative formula $$x _ { n + 1 } = \ln \left( 6 - x _ { n } \right) + 1 , \quad x _ { 0 } = 2$$ is used to find an approximate value for \(\alpha\).
2. Calculate the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\) to 4 decimal places.
3. By choosing a suitable interval, show that \(\alpha = 2.307\) correct to 3 decimal places.

1. $$f ( x ) = \sec x + 3 x - 2 , \quad - \frac { \pi } { 2 } < x < \frac { \pi } { 2 }$$

1. Show that there is a root of \(\mathrm { f } ( x ) = 0\) in the interval \([ 0.2,0.4 ]\)
2. Show that the equation \(\mathrm { f } ( x ) = 0\) can be written in the form $$x = \frac { 2 } { 3 } - \frac { 1 } { 3 \cos x }$$ The solution of \(\mathrm { f } ( x ) = 0\) is \(\alpha\), where \(\alpha = 0.3\) to 1 decimal place.
3. Starting with \(x _ { 0 } = 0.3\), use the iterative formula $$x _ { n + 1 } = \frac { 2 } { 3 } - \frac { 1 } { 3 \cos x _ { n } }$$ to calculate the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\), giving your answers to 4 decimal places.
4. State the value of \(\alpha\) correct to 3 decimal places.

4. $$f ( x ) = - x ^ { 3 } + 3 x ^ { 2 } - 1$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) can be rewritten as $$x = \sqrt { } \left( \frac { 1 } { 3 - x } \right)$$
2. Starting with \(x _ { 1 } = 0.6\), use the iteration $$\left. x _ { n + 1 } = \sqrt { ( } \frac { 1 } { 3 - x _ { n } } \right)$$ to calculate the values of \(x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving all your answers to 4 decimal places.
3. Show that \(x = 0.653\) is a root of \(\mathrm { f } ( x ) = 0\) correct to 3 decimal places.

7. $$f ( x ) = 3 x ^ { 3 } - 2 x - 6$$

1. Show that \(\mathrm { f } ( x ) = 0\) has a root, \(\alpha\), between \(x = 1.4\) and \(x = 1.45\)
2. Show that the equation \(\mathrm { f } ( x ) = 0\) can be written as $$x = \sqrt { } \left( \frac { 2 } { x } + \frac { 2 } { 3 } \right) , \quad x \neq 0$$
3. Starting with \(x _ { 0 } = 1.43\), use the iteration $$x _ { \mathrm { n } + 1 } = \sqrt { } \left( \frac { 2 } { x _ { \mathrm { n } } } + \frac { 2 } { 3 } \right)$$ to calculate the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\), giving your answers to 4 decimal places.
4. By choosing a suitable interval, show that \(\alpha = 1.435\) is correct to 3 decimal places.

1. \begin{figure}[h] \end{figure} Figure 1 shows part of the curve with equation \(y = - x ^ { 3 } + 2 x ^ { 2 } + 2\), which intersects the \(x\)-axis at the point \(A\) where \(x = \alpha\). To find an approximation to \(\alpha\), the iterative formula $$x _ { n + 1 } = \frac { 2 } { \left( x _ { n } \right) ^ { 2 } } + 2$$ is used.

1. Taking \(x _ { 0 } = 2.5\), find the values of \(x _ { 1 } , x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\). Give your answers to 3 decimal places where appropriate.
2. Show that \(\alpha = 2.359\) correct to 3 decimal places.

3. \(\mathrm { f } ( x ) = 4 \operatorname { cosec } x - 4 x + 1\), where \(x\) is in radians.

1. Show that there is a root \(\alpha\) of \(\mathrm { f } ( x ) = 0\) in the interval [1.2,1.3].
2. Show that the equation \(\mathrm { f } ( x ) = 0\) can be written in the form $$x = \frac { 1 } { \sin x } + \frac { 1 } { 4 }$$
3. Use the iterative formula $$x _ { n + 1 } = \frac { 1 } { \sin x _ { n } } + \frac { 1 } { 4 } , \quad x _ { 0 } = 1.25$$ to calculate the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\), giving your answers to 4 decimal places.
4. By considering the change of sign of \(\mathrm { f } ( x )\) in a suitable interval, verify that \(\alpha = 1.291\) correct to 3 decimal places.

2. $$f ( x ) = x ^ { 3 } + 3 x ^ { 2 } + 4 x - 12$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) can be written as $$x = \sqrt { } \left( \frac { 4 ( 3 - x ) } { ( 3 + x ) } \right) , \quad x \neq - 3$$ The equation \(x ^ { 3 } + 3 x ^ { 2 } + 4 x - 12 = 0\) has a single root which is between 1 and 2
2. Use the iteration formula $$x _ { n + 1 } = \sqrt { } \left( \frac { 4 \left( 3 - x _ { n } \right) } { \left( 3 + x _ { n } \right) } \right) , n \geqslant 0$$ with \(x _ { 0 } = 1\) to find, to 2 decimal places, the value of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\). The root of \(\mathrm { f } ( x ) = 0\) is \(\alpha\).
3. By choosing a suitable interval, prove that \(\alpha = 1.272\) to 3 decimal places.

6. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of part of the curve with equation $$y = 2 \cos \left( \frac { 1 } { 2 } x ^ { 2 } \right) + x ^ { 3 } - 3 x - 2$$ The curve crosses the \(x\)-axis at the point \(Q\) and has a minimum turning point at \(R\).

1. Show that the \(x\) coordinate of \(Q\) lies between 2.1 and 2.2
2. Show that the \(x\) coordinate of \(R\) is a solution of the equation $$x = \sqrt { 1 + \frac { 2 } { 3 } x \sin \left( \frac { 1 } { 2 } x ^ { 2 } \right) }$$ Using the iterative formula $$x _ { n + 1 } = \sqrt { 1 + \frac { 2 } { 3 } x _ { n } \sin \left( \frac { 1 } { 2 } x _ { n } ^ { 2 } \right) } , \quad x _ { 0 } = 1.3$$
3. find the values of \(x _ { 1 }\) and \(x _ { 2 }\) to 3 decimal places.

7. (a) Sketch the curve with equation \(y = \ln x\).
(b) Show that the tangent to the curve with equation \(y = \ln x\) at the point ( \(\mathrm { e } , 1\) ) passes through the origin.
(c) Use your sketch to explain why the line \(y = m x\) cuts the curve \(y = \ln x\) between \(x = 1\) and \(x = \mathrm { e }\) if \(0 < m < \frac { 1 } { \mathrm { e } }\). Taking \(x _ { 0 } = 1.86\) and using the iteration \(x _ { n + 1 } = \mathrm { e } ^ { \frac { 1 } { 3 } x _ { n } }\),
(d) calculate \(x _ { 1 } , x _ { 2 } , x _ { 3 } , x _ { 4 }\) and \(x _ { 5 }\), giving your answer to \(x _ { 5 }\) to 3 decimal places. The root of \(\ln x - \frac { 1 } { 3 } x = 0\) is \(\alpha\).
(e) By considering the change of sign of \(\ln x - \frac { 1 } { 3 } x\) over a suitable interval, show that your answer for \(x _ { 5 }\) is an accurate estimate of \(\alpha\), correct to 3 decimal places.

2. \(\quad \mathrm { f } ( x ) = x ^ { 3 } - 2 x - 5\).

1. Show that there is a root \(\alpha\) of \(\mathrm { f } ( x ) = 0\) for \(x\) in the interval \([ 2,3 ]\). The root \(\alpha\) is to be estimated using the iterative formula $$x _ { n + 1 } = \sqrt { \left( 2 + \frac { 5 } { x _ { n } } \right) } , \quad x _ { 0 } = 2$$
2. Calculate the values of \(x _ { 1 } , x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving your answers to 4 significant figures.
3. Prove that, to 5 significant figures, \(\alpha\) is 2.0946.

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.

2. $$f ( x ) = x ^ { 3 } - 3 x ^ { 2 } + \frac { 1 } { 2 \sqrt { x ^ { 5 } } } + 2 , \quad x > 0$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) in the interval \([ 2,3 ]\).
2. Taking 3 as a first approximation to \(\alpha\), apply the Newton-Raphson process once to \(\mathrm { f } ( x )\) to find a second approximation to \(\alpha\). Give your answer to 3 decimal places.

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

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) in the interval [1.6,1.7]
2. Taking 1.6 as a first approximation to \(\alpha\), apply the Newton-Raphson process once to \(\mathrm { f } ( x )\) to find a second approximation to \(\alpha\). Give your answer to 3 decimal places.

\(\mathrm { f } ( x ) = 2 ^ { x } - 10 \sin x - 2\), where \(x\) is measured in radians

1. Show that \(\mathrm { f } ( x ) = 0\) has a root, \(\alpha\), between 2 and 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.

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.
   [0pt]
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.

1. (a) Show that the equation \(4 x - 2 \sin x - 1 = 0\), where \(x\) is in radians, has a root \(\alpha\) in the interval [0.2, 0.6]
   [0pt] (b) Starting with the interval [0.2, 0.6], use interval bisection twice to find an interval of width 0.1 in which \(\alpha\) lies.
   (3)
   

4. $$f ( x ) = 1 - \frac { 1 } { 8 x ^ { 4 } } + \frac { 2 } { 7 \sqrt { x ^ { 7 } } } \quad x > 0$$ The equation \(\mathrm { f } ( x ) = 0\) has a single root, \(\alpha\), that lies in the interval \([ 0.15,0.25 ]\)

1. 1. Determine \(\mathrm { f } ^ { \prime } ( x )\)
   2. Explain why 0.25 cannot be used as an initial approximation for \(\alpha\) in the Newton-Raphson process.
   3. Taking 0.15 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. Use linear interpolation once on the interval \([ 0.15,0.25 ]\) 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]
   [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.