Apply iteration to find root

2 Let \(\mathrm { f } ( x ) = 2 x ^ { 3 } - 5 x ^ { 2 } + 4\).

1. Show that if a sequence of values given by the iterative formula $$x _ { n + 1 } = \sqrt { \frac { 4 } { 5 - 2 x _ { n } } }$$ converges, then it converges to a root of the equation \(\mathrm { f } ( x ) = 0\).
2. The equation has a root close to 1.2 . Use the iterative formula from part (a) and an initial value of 1.2 to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

7. \begin{figure}[h] \end{figure} Figure 3 shows a sketch of part of the curve with equation $$y = 2 \cos 3 x - 3 x + 4 \quad x > 0$$ where \(x\) is measured in radians. The curve crosses the \(x\)-axis at the point \(P\), as shown in Figure 3.
Given that the \(x\) coordinate of \(P\) is \(\alpha\),

1. show that \(\alpha\) lies between 0.8 and 0.9 The iteration formula $$x _ { n + 1 } = \frac { 1 } { 3 } \arccos \left( 1.5 x _ { n } - 2 \right)$$ can be used to find an approximate value for \(\alpha\).
2. Using this iteration formula with \(x _ { 1 } = 0.8\) find, to 4 decimal places, the value of
   1. \(X _ { 2 }\)
   2. \(X _ { 5 }\) The point \(Q\) and the point \(R\) are local minimum points on the curve, as shown in Figure 3.
      Given that the \(x\) coordinates of \(Q\) and \(R\) are \(\beta\) and \(\lambda\) respectively, and that they are the two smallest values of \(x\) at which local minima occur,
3. find, using calculus, the exact value of \(\beta\) and the exact value of \(\lambda\).

1. A curve has equation \(y = \mathrm { f } ( x )\) where

$$\mathrm { f } ( x ) = x ^ { 4 } - 5 x ^ { 2 } + 4 x - 7 \quad x \in \mathbb { R }$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root, \(\alpha\), in the interval [2,3]
2. Show that the equation \(\mathrm { f } ( x ) = 0\) can be written as $$x = \sqrt [ 3 ] { \frac { 5 x ^ { 2 } - 4 x + 7 } { x } }$$ The iterative formula $$x _ { n + 1 } = \sqrt [ 3 ] { \frac { 5 x _ { n } ^ { 2 } - 4 x _ { n } + 7 } { x _ { n } } }$$ is used to find \(\alpha\)
3. Starting with \(x _ { 1 } = 2\) and using the iterative formula,
   1. find, to 4 decimal places, the value of \(x _ { 2 }\)
   2. find, to 4 decimal places, the value of \(\alpha\)

1. $$g ( x ) = x ^ { 6 } + 2 x - 1000$$

1. Show that \(\mathrm { g } ( x ) = 0\) has a root \(\alpha\) in the interval [3,4] Using the iteration formula $$x _ { n + 1 } = \sqrt [ 6 ] { 1000 - 2 x _ { n } } \quad \text { with } x _ { 1 } = 3$$
2. 1. find, to 4 decimal places, the value of \(x _ { 2 }\)
   2. find, by repeated iteration, the value of \(\alpha\). Give your answer to 4 decimal places.

1. A curve has equation \(y = \mathrm { f } ( x )\) where

$$\mathrm { f } ( x ) = x ^ { 2 } - 5 x + \mathrm { e } ^ { x } \quad x \in \mathbb { R }$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root, \(\alpha\), in the interval [1,2] The iterative formula $$x _ { n + 1 } = \sqrt { 5 x _ { n } - \mathrm { e } ^ { x _ { n } } }$$ with \(x _ { 1 } = 1\) is used to find an approximate value for the root \(\alpha\).
2. 1. Find the value of \(x _ { 2 }\) to 4 decimal places.
   2. Find, by repeated iteration, the value of \(\alpha\), giving your answer to 4 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 $$x _ { n + 1 } = \sqrt { \left( \frac { 4 \left( 3 - x _ { n } \right) } { \left( 3 + x _ { n } \right) } \right) } \quad n \geqslant 0$$ 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.
   

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5. $$f ( x ) = - x ^ { 3 } + 4 x ^ { 2 } - 6$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root between \(x = 1\) and \(x = 2\)
2. Show that the equation \(\mathrm { f } ( x ) = 0\) can be rewritten as $$x = \sqrt { \left( \frac { 6 } { 4 - x } \right) }$$
3. Starting with \(x _ { 1 } = 1.5\) use the iteration \(x _ { n + 1 } = \sqrt { \left( \frac { 6 } { 4 - x _ { n } } \right) }\) to calculate the values of \(x _ { 2 }\), \(x _ { 3 }\) and \(x _ { 4 }\) giving all your answers to 4 decimal places.
4. Using a suitable interval, show that 1.572 is a root of \(\mathrm { f } ( x ) = 0\) correct to 3 decimal places.

2. $$f ( x ) = x ^ { 3 } - 5 x + 16$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) can be rewritten as $$x = ( a x + b ) ^ { \frac { 1 } { 3 } }$$ giving the values of the constants \(a\) and \(b\). The equation \(\mathrm { f } ( x ) = 0\) has exactly one real root \(\alpha\), where \(\alpha = - 3\) to one significant figure.
2. Starting with \(x _ { 1 } = - 3\), use the iteration $$x _ { n + 1 } = \left( a x _ { n } + b \right) ^ { \frac { 1 } { 3 } }$$ with the values of \(a\) and \(b\) found in part (a), to calculate the values of \(x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving all your answers to 3 decimal places.
3. Using a suitable interval, show that \(\alpha = - 3.17\) correct to 2 decimal places.

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

1. Show that the equation \(\mathrm { f } ( x ) = 0\) can be rewritten as $$x = \frac { 1 } { 2 } \mathrm { e } ^ { - \frac { 1 } { 4 } x ^ { 2 } }$$ The equation \(\mathrm { f } ( x ) = 0\) has a root near 0.5
2. Starting with \(x _ { 1 } = 0.5\) use the iterative formula $$x _ { n + 1 } = \frac { 1 } { 2 } \mathrm { e } ^ { - \frac { 1 } { 4 } x _ { n } ^ { 2 } }$$ to calculate the values of \(x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving your answers to 4 decimal places.
3. Using a suitable interval, show that 0.473 is a root of \(\mathrm { f } ( x ) = 0\) correct to 3 decimal places.

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. $$f ( x ) = 2 x ^ { 3 } + x - 20$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) can be rewritten as $$x = \sqrt [ 3 ] { a - b x }$$ where \(a\) and \(b\) are positive constants to be determined.
2. Starting with \(x _ { 1 } = 2.1\) use the iteration formula \(x _ { n + 1 } = \sqrt [ 3 ] { a - b x _ { n } }\), with the numerical values of \(a\) and \(b\), to calculate the values of \(x _ { 2 }\) and \(x _ { 3 }\) giving your answers to 3 decimal places.
3. Using a suitable interval, show that 2.077 is a root of the equation \(\mathrm { f } ( x ) = 0\) correct to 3 decimal places.
4. Hence state a root, to 3 decimal places, of the equation $$2 ( x + 2 ) ^ { 3 } + x - 18 = 0$$

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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.

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.

7. $$f ( x ) = 3 x e ^ { x } - 1$$ The curve with equation \(y = \mathrm { f } ( x )\) has a turning point \(P\).

1. Find the exact coordinates of \(P\). The equation \(\mathrm { f } ( x ) = 0\) has a root between \(x = 0.25\) and \(x = 0.3\)
2. Use the iterative formula $$x _ { n + 1 } = \frac { 1 } { 3 } \mathrm { e } ^ { - x _ { n } }$$ with \(x _ { 0 } = 0.25\) to find, to 4 decimal places, the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\).
3. By choosing a suitable interval, show that a root of \(\mathrm { f } ( x ) = 0\) is \(x = 0.2576\) correct to 4 decimal places.

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

1. Show that \(\mathrm { f } ( x ) = 0\) can be rearranged as $$x = \sqrt { } \left( \frac { 3 x + 11 } { x + 2 } \right) , \quad x \neq - 2 .$$ The equation \(\mathrm { f } ( x ) = 0\) has one positive root \(\alpha\). The iterative formula \(x _ { n + 1 } = \sqrt { } \left( \frac { 3 x _ { n } + 11 } { x _ { n } + 2 } \right)\) is used to find an approximation to \(\alpha\).
2. Taking \(x _ { 1 } = 0\), find, to 3 decimal places, the values of \(x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\).
3. Show that \(\alpha = 2.057\) 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.

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.

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.

3

1. It is given that \(a\) and \(b\) are positive constants. By sketching graphs of $$y = x ^ { 5 } \quad \text { and } \quad y = a - b x$$ on the same diagram, show that the equation $$x ^ { 5 } + b x - a = 0$$ has exactly one real root.
2. Use the iterative formula \(x _ { n + 1 } = \sqrt [ 5 ] { 53 - 2 x _ { n } }\), with a suitable starting value, to find the real root of the equation \(x ^ { 5 } + 2 x - 53 = 0\). Show the result of each iteration, and give the root correct to 3 decimal places.

3 The equation \(2 x ^ { 3 } + 4 x - 35 = 0\) has one real root.

1. Show by calculation that this real root lies between 2 and 3 .
2. Use the iterative formula $$x _ { n + 1 } = \sqrt [ 3 ] { 17.5 - 2 x _ { n } }$$ with a suitable starting value, to find the real root of the equation \(2 x ^ { 3 } + 4 x - 35 = 0\) correct to 2 decimal places. You should show the result of each iteration.

3 The sequence defined by the iterative formula $$x _ { n + 1 } = \sqrt [ 3 ] { } \left( 17 - 5 x _ { n } \right)$$ with \(x _ { 1 } = 2\), converges to \(\alpha\).

1. Use the iterative formula to find \(\alpha\) correct to 2 decimal places. You should show the result of each iteration.
2. Find a cubic equation of the form $$x ^ { 3 } + c x + d = 0$$ which has \(\alpha\) as a root.
3. Does this cubic equation have any other real roots? Justify your answer.