1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams

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.

1. The function \(f\) is defined by

$$\mathrm { f } : x \mapsto \ln ( 4 - 2 x ) , \quad x < 2 \quad \text { and } \quad x \in \mathbb { R } .$$

1. Show that the inverse function of f is defined by $$\mathrm { f } ^ { - 1 } : x \mapsto 2 - \frac { 1 } { 2 } \mathrm { e } ^ { x }$$ and write down the domain of \(\mathrm { f } ^ { - 1 }\).
2. Write down the range of \(\mathrm { f } ^ { - 1 }\).
3. In the space provided on page 16, sketch the graph of \(y = f ^ { - 1 } ( x )\). State the coordinates of the points of intersection with the \(x\) and \(y\) axes. The graph of \(y = x + 2\) crosses the graph of \(y = f ^ { - 1 } ( x )\) at \(x = k\). The iterative formula $$x _ { n + 1 } = - \frac { 1 } { 2 } e ^ { x _ { n } } , x _ { 0 } = - 0.3$$ is used to find an approximate value for \(k\).
4. Calculate the values of \(x _ { 1 }\) and \(x _ { 2 }\), giving your answers to 4 decimal places.
5. Find the value of \(k\) 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.

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. $$\mathrm { f } ( x ) = 3 \mathrm { e } ^ { x } - \frac { 1 } { 2 } \ln x - 2 , \quad x > 0 .$$

1. Differentiate to find \(\mathrm { f } ^ { \prime } ( x )\). The curve with equation \(y = \mathrm { f } ( x )\) has a turning point at \(P\). The \(x\)-coordinate of \(P\) is \(\alpha\).
2. Show that \(\alpha = \frac { 1 } { 6 } \mathrm { e } ^ { - \alpha }\). The iterative formula $$x _ { n + 1 } = \frac { 1 } { 6 } \mathrm { e } ^ { - x _ { n } } , x _ { 0 } = 1$$ is used to find an approximate value for \(\alpha\).
3. Calculate the values of \(x _ { 1 } , x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving your answers to 4 decimal places.
4. By considering the change of sign of \(\mathrm { f } ^ { \prime } ( x )\) in a suitable interval, prove that \(\alpha = 0.1443\) correct to 4 decimal places.

5. \begin{figure}[h] \end{figure} Figure 2 shows part of the curve with equation $$y = ( 2 x - 1 ) \tan 2 x , \quad 0 \leqslant x < \frac { \pi } { 4 }$$ The curve has a minimum at the point \(P\). The \(x\)-coordinate of \(P\) is \(k\).

1. Show that \(k\) satisfies the equation $$4 k + \sin 4 k - 2 = 0$$ The iterative formula $$x _ { n + 1 } = \frac { 1 } { 4 } \left( 2 - \sin 4 x _ { n } \right) , x _ { 0 } = 0.3$$ is used to find an approximate value for \(k\).
2. Calculate the values of \(x _ { 1 } , x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving your answers to 4 decimal places.
3. Show that \(k = 0.277\), correct to 3 significant figures.

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.

7. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of part of the curve with equation \(y = \mathrm { f } ( x )\) where $$\mathrm { f } ( x ) = \left( x ^ { 2 } + 3 x + 1 \right) \mathrm { e } ^ { x ^ { 2 } }$$ The curve cuts the \(x\)-axis at points \(A\) and \(B\) as shown in Figure 2 .

1. Calculate the \(x\) coordinate of \(A\) and the \(x\) coordinate of \(B\), giving your answers to 3 decimal places.
2. Find \(\mathrm { f } ^ { \prime } ( x )\). The curve has a minimum turning point at the point \(P\) as shown in Figure 2.
3. Show that the \(x\) coordinate of \(P\) is the solution of $$x = - \frac { 3 \left( 2 x ^ { 2 } + 1 \right) } { 2 \left( x ^ { 2 } + 2 \right) }$$
4. Use the iteration formula $$x _ { n + 1 } = - \frac { 3 \left( 2 x _ { n } ^ { 2 } + 1 \right) } { 2 \left( x _ { n } ^ { 2 } + 2 \right) } , \quad \text { with } x _ { 0 } = - 2.4$$ to calculate the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\), giving your answers to 3 decimal places. The \(x\) coordinate of \(P\) is \(\alpha\).
5. By choosing a suitable interval, prove that \(\alpha = - 2.43\) to 2 decimal places.

4. $$\mathrm { f } ( x ) = 25 x ^ { 2 } \mathrm { e } ^ { 2 x } - 16 , \quad x \in \mathbb { R }$$

1. Using calculus, find the exact coordinates of the turning points on the curve with equation \(y = \mathrm { f } ( x )\).
2. Show that the equation \(\mathrm { f } ( x ) = 0\) can be written as \(x = \pm \frac { 4 } { 5 } \mathrm { e } ^ { - x }\) The equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\), where \(\alpha = 0.5\) to 1 decimal place.
3. Starting with \(x _ { 0 } = 0.5\), use the iteration formula $$x _ { n + 1 } = \frac { 4 } { 5 } \mathrm { e } ^ { - x _ { n } }$$ to calculate the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\), giving your answers to 3 decimal places.
4. Give an accurate estimate for \(\alpha\) to 2 decimal places, and justify your answer.

2. A curve \(C\) has equation \(y = \mathrm { e } ^ { 4 x } + x ^ { 4 } + 8 x + 5\)

1. Show that the \(x\) coordinate of any turning point of \(C\) satisfies the equation $$x ^ { 3 } = - 2 - \mathrm { e } ^ { 4 x }$$
2. On the axes given on page 5, sketch, on a single diagram, the curves with equations
   1. \(y = x ^ { 3 }\),
   2. \(y = - 2 - e ^ { 4 x }\) On your diagram give the coordinates of the points where each curve crosses the \(y\)-axis and state the equation of any asymptotes.
3. Explain how your diagram illustrates that the equation \(x ^ { 3 } = - 2 - e ^ { 4 x }\) has only one root. The iteration formula $$x _ { n + 1 } = \left( - 2 - \mathrm { e } ^ { 4 x _ { n } } \right) ^ { \frac { 1 } { 3 } } , \quad x _ { 0 } = - 1$$ can be used to find an approximate value for this root.
4. Calculate the values of \(x _ { 1 }\) and \(x _ { 2 }\), giving your answers to 5 decimal places.
5. Hence deduce the coordinates, to 2 decimal places, of the turning point of the curve \(C\). \includegraphics[max width=\textwidth, alt={}, center]{be00fdaa-2fe3-4f06-a710-08ec67fb911e-04_1285_1294_308_331}

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.

6. \begin{figure}[h] \end{figure} Figure 1 is a sketch showing part of the curve with equation \(y = 2 ^ { x + 1 } - 3\) and part of the line with equation \(y = 17 - x\). The curve and the line intersect at the point \(A\).

1. Show that the \(x\) coordinate of \(A\) satisfies the equation $$x = \frac { \ln ( 20 - x ) } { \ln 2 } - 1$$
2. Use the iterative formula $$x _ { n + 1 } = \frac { \ln \left( 20 - x _ { n } \right) } { \ln 2 } - 1 , \quad x _ { 0 } = 3$$ to calculate the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\), giving your answers to 3 decimal places.
3. Use your answer to part (b) to deduce the coordinates of the point \(A\), giving your answers to one decimal place.

4. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the curve with equation \(y = g ( x )\), where $$\mathrm { g } ( x ) = \left| 4 \mathrm { e } ^ { 2 x } - 25 \right| , \quad x \in \mathbb { R }$$ The curve cuts the \(y\)-axis at the point \(A\) and meets the \(x\)-axis at the point \(B\). The curve has an asymptote \(y = k\), where \(k\) is a constant, as shown in Figure 1

1. Find, giving each answer in its simplest form,
   1. the \(y\) coordinate of the point \(A\),
   2. the exact \(x\) coordinate of the point \(B\),
   3. the value of the constant \(k\). The equation \(\mathrm { g } ( x ) = 2 x + 43\) has a positive root at \(x = \alpha\)
2. Show that \(\alpha\) is a solution of \(x = \frac { 1 } { 2 } \ln \left( \frac { 1 } { 2 } x + 17 \right)\) The iteration formula $$x _ { n + 1 } = \frac { 1 } { 2 } \ln \left( \frac { 1 } { 2 } x _ { n } + 17 \right)$$ can be used to find an approximation for \(\alpha\)
3. Taking \(x _ { 0 } = 1.4\) find the values of \(x _ { 1 }\) and \(x _ { 2 }\) Give each answer to 4 decimal places.
4. By choosing a suitable interval, show that \(\alpha = 1.437\) to 3 decimal places. \includegraphics[max width=\textwidth, alt={}, center]{d3ba2776-eedb-48f0-834f-41aa454afba3-07_2258_47_315_37}

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.