Rearrange to iterative form

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

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 $$2 ( x + 2 ) ^ { 3 } + x - 18 = 0$$

   VIIIV SIHI NI JIIYM ION OC

   VIIV SIHI NI JIIIM ION OC

   VIIV SIHI NI JIIYM ION OC

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.

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.

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.

5. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of part of the curve \(C\) with equation $$y = 2 \ln ( 2 x + 5 ) - \frac { 3 x } { 2 } , \quad x > - 2.5$$ The point \(P\) with \(x\) coordinate - 2 lies on \(C\).

1. Find an equation of the normal to \(C\) at \(P\). Write your answer in the form \(a x + b y = c\), where \(a\), \(b\) and \(c\) are integers. The normal to \(C\) at \(P\) cuts the curve again at the point \(Q\), as shown in Figure 2
2. Show that the \(x\) coordinate of \(Q\) is a solution of the equation $$x = \frac { 20 } { 11 } \ln ( 2 x + 5 ) - 2$$ The iteration formula $$x _ { n + 1 } = \frac { 20 } { 11 } \ln \left( 2 x _ { n } + 5 \right) - 2$$ can be used to find an approximation for the \(x\) coordinate of \(Q\).
3. Taking \(x _ { 1 } = 2\), find the values of \(x _ { 2 }\) and \(x _ { 3 }\), giving each answer to 4 decimal places.

4. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the curve \(C\) with equation $$y = \mathrm { e } ^ { - 2 x } + x ^ { 2 } - 3$$ The curve \(C\) crosses the \(y\)-axis at the point \(A\). The line \(l\) is the normal to \(C\) at the point \(A\).

1. Find the equation of \(l\), writing your answer in the form \(y = m x + c\), where \(m\) and \(c\) are constants. The line \(l\) meets \(C\) again at the point \(B\), as shown in Figure 1 .
2. Show that the \(x\) coordinate of \(B\) is a solution of $$x = \sqrt { 1 + \frac { 1 } { 2 } x - \mathrm { e } ^ { - 2 x } }$$ Using the iterative formula $$x _ { n + 1 } = \sqrt { 1 + \frac { 1 } { 2 } x _ { n } - \mathrm { e } ^ { - 2 x _ { n } } }$$ with \(x _ { 1 } = 1\)
3. find \(x _ { 2 }\) and \(x _ { 3 }\) to 3 decimal places.

8. The population in thousands, \(P\), of a town at time \(t\) years after \(1 ^ { \text {st } }\) January 1980 is modelled by the formula $$P = 30 + 50 \mathrm { e } ^ { 0.002 t }$$ Use this model to estimate

1. the population of the town on \(1 ^ { \text {st } }\) January 2010,
2. the year in which the population first exceeds 84000 . The population in thousands, \(Q\), of another town is modelled by the formula $$Q = 26 + 50 \mathrm { e } ^ { 0.003 t }$$
3. Show that the value of \(t\) when \(P = Q\) is a solution of the equation $$t = 1000 \ln \left( 1 + 0.08 \mathrm { e } ^ { - 0.002 t } \right)$$
4. Use the iterative formula $$t _ { n + 1 } = 1000 \ln \left( 1 + 0.08 \mathrm { e } ^ { - 0.002 t _ { n } } \right)$$ with \(t _ { 0 } = 50\) to find \(t _ { 1 } , t _ { 2 }\) and \(t _ { 3 }\) and hence, the year in which the populations of these two towns will be equal according to these models.

5. \(\mathrm { f } ( x ) = \frac { 1 } { 3 } x ^ { 3 } - 4 x - 2\) a. Show that the equation \(\mathrm { f } ( x ) = 0\) can be written in the form \(x = \pm \sqrt { a + \frac { b } { x } }\), and state the values of the integers \(a\) and \(b\). \(\mathrm { f } ( x ) = 0\) has one positive root, \(\alpha\).
The iterative formula \(x _ { n + 1 } = \sqrt { a + \frac { b } { x _ { n } } } , \quad x _ { 0 } = 4\) is used to find an approximation value for \(\alpha\).
b. Calculate the values of \(x _ { 1 } , x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\) to 4 decimal places.
c. Explain why for this question, the Newton-Raphson method cannot be used with \(x _ { 1 } = 2\).

3 The equation $$x + ( 1 + 3 x ) ^ { \frac { 1 } { 4 } } = 0$$ has a single root, \(\alpha\).

1. Show that \(\alpha\) lies between - 0.33 and - 0.32 .
2. Show that the equation \(x + ( 1 + 3 x ) ^ { \frac { 1 } { 4 } } = 0\) can be rearranged into the form $$x = \frac { 1 } { 3 } \left( x ^ { 4 } - 1 \right)$$
3. Use the iteration \(x _ { n + 1 } = \frac { \left( x _ { n } ^ { 4 } - 1 \right) } { 3 }\) with \(x _ { 1 } = - 0.3\) to find \(x _ { 4 }\), giving your answer to three significant figures.

2 A curve is defined by the equation \(y = \left( x ^ { 2 } - 4 \right) \ln ( x + 2 )\) for \(x \geqslant 3\).
The curve intersects the line \(y = 15\) at a single point, where \(x = \alpha\).

1. Show that \(\alpha\) lies between 3.5 and 3.6.
2. Show that the equation \(\left( x ^ { 2 } - 4 \right) \ln ( x + 2 ) = 15\) can be arranged into the form $$x = \pm \sqrt { 4 + \frac { 15 } { \ln ( x + 2 ) } }$$ (2 marks)
3. Use the iteration $$x _ { n + 1 } = \sqrt { 4 + \frac { 15 } { \ln \left( x _ { n } + 2 \right) } }$$ with \(x _ { 1 } = 3.5\) to find the values of \(x _ { 2 }\) and \(x _ { 3 }\), giving your answers to three decimal places.
   (2 marks)

1

1. Show that the equation \(x ^ { 3 } - 6 x + 1 = 0\) has a root \(\alpha\), where \(2 < \alpha < 3\).
2. Show that the equation \(x ^ { 3 } - 6 x + 1 = 0\) can be rearranged into the form $$x ^ { 2 } = 6 - \frac { 1 } { x }$$ (1 mark)
3. Use the recurrence relation \(x _ { n + 1 } = \sqrt { 6 - \frac { 1 } { x _ { n } } }\), with \(x _ { 1 } = 2.5\), to find the value of \(x _ { 3 }\), giving your answer to four significant figures.
   (2 marks)

   \includegraphics[max width=\textwidth, alt={}]{b8614dd6-2197-40c3-a673-5bef3e3653a5-2_142_116_2560_157}

   \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\)

1 The curve \(y = x ^ { 3 } - x - 7\) intersects the \(x\)-axis at the point where \(x = \alpha\).

1. Show that \(\alpha\) lies between 2.0 and 2.1.
2. Show that the equation \(x ^ { 3 } - x - 7 = 0\) can be rearranged in the form \(x = \sqrt [ 3 ] { x + 7 }\).
3. Use the iteration \(x _ { n + 1 } = \sqrt [ 3 ] { x _ { n } + 7 }\) with \(x _ { 1 } = 2\) to find the values of \(x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving your answers to three significant figures.

2 The table shows some values of \(x\) and the associated values of \(y = f ( x )\).

1. Calculate an estimate of \(\frac { \mathrm { dy } } { \mathrm { dx } }\) at \(x = 3\) using the forward difference method, giving your answer correct to \(\mathbf { 5 }\) decimal places.
2. Calculate an estimate of \(\frac { \mathrm { dy } } { \mathrm { dx } }\) at \(x = 3\) using the central difference method, giving your answer correct to \(\mathbf { 5 }\) decimal places.
3. Explain why your answer to part (b) is likely to be closer than your answer to part (a) to the true value of \(\frac { \mathrm { dy } } { \mathrm { dx } }\) at \(x = 3\). When \(x = 5\) it is given that \(y = 1.4645\) and \(\frac { \mathrm { dy } } { \mathrm { dx } } = 0.1820\), correct to 4 decimal places.
4. Determine an estimate of the error when \(\mathrm { f } ( 5 )\) is used to estimate \(\mathrm { f } ( 5.024 )\).

6 [Figure 1, printed on the insert, is provided for use in this question.]
The curve \(y = x ^ { 3 } + 4 x - 3\) intersects the \(x\)-axis at the point \(A\) where \(x = \alpha\).

1. Show that \(\alpha\) lies between 0.5 and 1.0.
2. Show that the equation \(x ^ { 3 } + 4 x - 3 = 0\) can be rearranged into the form \(x = \frac { 3 - x ^ { 3 } } { 4 }\).
   (1 mark)
3. 1. Use the iteration \(x _ { n + 1 } = \frac { 3 - x _ { n } { } ^ { 3 } } { 4 }\) with \(x _ { 1 } = 0.5\) to find \(x _ { 3 }\), giving your answer to two decimal places.
   2. The sketch on Figure 1 shows parts of the graphs of \(y = \frac { 3 - x ^ { 3 } } { 4 }\) and \(y = x\), and the position of \(x _ { 1 }\). On Figure 1, draw a cobweb or staircase diagram to show how convergence takes place, indicating the positions of \(x _ { 2 }\) and \(x _ { 3 }\) on the \(x\)-axis.
      (3 marks)

6 [Figure 1, printed on the insert, is provided for use in this question.]
The curve \(y = x ^ { 3 } + 4 x - 3\) intersects the \(x\)-axis at the point \(A\) where \(x = \alpha\).

1. Show that \(\alpha\) lies between 0.5 and 1.0.
2. Show that the equation \(x ^ { 3 } + 4 x - 3 = 0\) can be rearranged into the form \(x = \frac { 3 - x ^ { 3 } } { 4 }\).
   (1 mark)
3. 1. Use the iteration \(x _ { n + 1 } = \frac { 3 - x _ { n } { } ^ { 3 } } { 4 }\) with \(x _ { 1 } = 0.5\) to find \(x _ { 3 }\), giving your answer to two decimal places.
      (3 marks)
   2. The sketch on Figure 1 shows parts of the graphs of \(y = \frac { 3 - x ^ { 3 } } { 4 }\) and \(y = x\), and the position of \(x _ { 1 }\). On Figure 1, draw a cobweb or staircase diagram to show how convergence takes place, indicating the positions of \(x _ { 2 }\) and \(x _ { 3 }\) on the \(x\)-axis.
      (3 marks)

3 [Figure 1, printed on the insert, is provided for use in this question.]
The curve with equation \(y = x ^ { 3 } + 5 x - 4\) intersects the \(x\)-axis at the point \(A\), where \(x = \alpha\).

1. Show that \(\alpha\) lies between 0.5 and 1 .
2. Show that the equation \(x ^ { 3 } + 5 x - 4 = 0\) can be rearranged into the form $$x = \frac { 1 } { 5 } \left( 4 - x ^ { 3 } \right)$$
3. Use the iteration \(x _ { n + 1 } = \frac { 1 } { 5 } \left( 4 - x _ { n } { } ^ { 3 } \right)\) with \(x _ { 1 } = 0.5\) to find \(x _ { 3 }\), giving your answer to three decimal places.
4. The sketch on Figure 1 shows parts of the graphs of \(y = \frac { 1 } { 5 } \left( 4 - x ^ { 3 } \right)\) and \(y = x\), and the position of \(x _ { 1 }\). On Figure 1, draw a cobweb or staircase diagram to show how convergence takes place, indicating the positions of \(x _ { 2 }\) and \(x _ { 3 }\) on the \(x\)-axis.

11 The daily world production of oil can be modelled using $$V = 10 + 100 \left( \frac { t } { 30 } \right) ^ { 3 } - 50 \left( \frac { t } { 30 } \right) ^ { 4 }$$ where \(V\) is volume of oil in millions of barrels, and \(t\) is time in years since 1 January 1980. 11

1. 1. The model is used to predict the time, \(T\), when oil production will fall to zero.
      Show that \(T\) satisfies the equation $$T = \sqrt [ 3 ] { 60 T ^ { 2 } + \frac { 162000 } { T } }$$ 11
      1. (ii) Use the iterative formula \(T _ { n + 1 } = \sqrt [ 3 ] { 60 T _ { n } { } ^ { 2 } + \frac { 162000 } { T _ { n } } }\), with \(T _ { 0 } = 38\), to find the values of \(T _ { 1 } , T _ { 2 }\), and \(T _ { 3 }\), giving your answers to three decimal places.
         11
   2. (iii) Explain the relevance of using \(T _ { 0 } = 38\) 11
   3. From 1 January 1980 the daily use of oil by one technologically developing country can be modelled as $$V = 4.5 \times 1.063 ^ { t }$$ Use the models to show that the country's use of oil and the world production of oil will be equal during the year 2029.
      [0pt] [4 marks] \(12 \quad \mathrm { p } ( x ) = 30 x ^ { 3 } - 7 x ^ { 2 } - 7 x + 2\)

\includegraphics{figure_6} The diagram shows the curve with equation \(y = \frac{\ln(2x + 1)}{x + 3}\). The curve has a maximum point M.

1. Find an expression for \(\frac{dy}{dx}\). [2]
2. Show that the x-coordinate of M satisfies the equation \(x = \frac{x + 3}{\ln(2x + 1)} - 0.5\). [2]
3. Show by calculation that the x-coordinate of M lies between 2.5 and 3.0. [2]
4. Use an iterative formula based on the equation in part (b) to find the x-coordinate of M correct to 4 significant figures. Give the result of each iteration to 6 significant figures. [3]

\includegraphics{figure_4} The diagram shows the curve with equation $$y = x^4 + 2x^3 + 2x^2 - 12x - 32.$$ The curve crosses the \(x\)-axis at points with coordinates \((\alpha, 0)\) and \((\beta, 0)\).

1. Use the factor theorem to show that \((x + 2)\) is a factor of $$x^4 + 2x^3 + 2x^2 - 12x - 32.$$ [2]
2. Show that \(\beta\) satisfies an equation of the form \(x = \sqrt[3]{p + qx}\), and state the values of \(p\) and \(q\). [3]
3. Use an iterative formula based on the equation in part (ii) to find the value of \(\beta\) correct to 4 significant figures. Give the result of each iteration to 6 significant figures. [3]

\includegraphics{figure_2} Figure 2 shows part of the curve \(C\) with equation \(y = f(x)\), where $$f(x) = 0.5e^x - x^2.$$ The curve \(C\) cuts the \(y\)-axis at \(A\) and there is a minimum at the point \(B\).

1. Find an equation of the tangent to \(C\) at \(A\). [4]

The \(x\)-coordinate of \(B\) is approximately \(2.15\). A more exact estimate is to be made of this coordinate using iterations \(x_{n+1} = \ln g(x_n)\).

1. Show that a possible form for \(g(x)\) is \(g(x) = 4x\). [3]
2. Using \(x_{n+1} = \ln 4x_n\), with \(x_0 = 2.15\), calculate \(x_1\), \(x_2\) and \(x_3\). Give the value of \(x_3\) to 4 decimal places. [2]