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

6. $$\mathrm { f } ( x ) = x \cos \left( \frac { x } { 3 } \right) \quad x > 0$$

1. Find \(\mathrm { f } ^ { \prime } ( x )\)
2. Show that the equation \(\mathrm { f } ^ { \prime } ( x ) = 0\) can be written as $$x = k \arctan \left( \frac { k } { x } \right)$$ where \(k\) is an integer to be found.
3. Starting with \(x _ { 1 } = 2.5\) use the iteration formula $$x _ { n + 1 } = k \arctan \left( \frac { k } { x _ { n } } \right)$$ with the value of \(k\) found in part (b), to calculate the values of \(x _ { 2 }\) and \(x _ { 6 }\) giving your answers to 3 decimal places.
4. Using a suitable interval and a suitable function that should be stated, show that a root of \(\mathrm { f } ^ { \prime } ( x ) = 0\) is 2.581 correct to 3 decimal places.
   In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.

5. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the curve with equation $$y = 6 \ln ( 2 x + 3 ) - \frac { 1 } { 2 } x ^ { 2 } + 4 \quad x > - \frac { 3 } { 2 }$$ The curve cuts the negative \(x\)-axis at the point \(P\), as shown in Figure 1.

1. Show that the \(x\) coordinate of \(P\) lies in the interval \([ - 1.25 , - 1.2 ]\) The curve cuts the positive \(x\)-axis at the point \(Q\), also shown in Figure 1.
   Using the iterative formula $$x _ { n + 1 } = \sqrt { 12 \ln \left( 2 x _ { n } + 3 \right) + 8 } \text { with } x _ { 1 } = 6$$
2. 1. find, to 4 decimal places, the value of \(x _ { 2 }\)
   2. find, by continued iteration, the \(x\) coordinate of \(Q\). Give your answer to 4 decimal places. The curve has a maximum turning point at \(M\), as shown in Figure 1.
3. Using calculus and showing each stage of your working, find the \(x\) coordinate of \(M\).

9. \begin{figure}[h] \end{figure} Figure 3 shows a sketch of part of the curve \(C\) with equation $$y = \sqrt { 3 + 4 \mathrm { e } ^ { x ^ { 2 } } } \quad x \geqslant 0$$

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\), giving your answer in simplest form. The point \(P\) with \(x\) coordinate \(\alpha\) lies on \(C\).
   Given that the tangent to \(C\) at \(P\) passes through the origin, as shown in Figure 3,
2. show that \(x = \alpha\) is a solution of the equation $$4 x ^ { 2 } e ^ { x ^ { 2 } } - 4 e ^ { x ^ { 2 } } - 3 = 0$$
3. Hence show that \(\alpha\) lies between 1 and 2
4. Show that the equation in part (b) can be written in the form $$x = \frac { 1 } { 2 } \sqrt { 4 + 3 \mathrm { e } ^ { - x ^ { 2 } } }$$ The iteration formula $$x _ { n + 1 } = \frac { 1 } { 2 } \sqrt { 4 + 3 \mathrm { e } ^ { - x _ { n } ^ { 2 } } }$$ with \(x _ { 1 } = 1\) is used to find an approximation for \(\alpha\).
5. Use the iteration formula to find, to 4 decimal places, the value of
   1. \(X _ { 3 }\)
   2. \(\alpha\)

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. The curve \(C\) has equation

$$y = x ^ { 2 } \cos \left( \frac { 1 } { 2 } x \right) \quad 0 < x \leqslant \pi$$ The curve has a stationary point at the point \(P\).

1. Show, using calculus, that the \(x\) coordinate of \(P\) is a solution of the equation $$x = 2 \arctan \left( \frac { 4 } { x } \right)$$ Using the iteration formula $$x _ { n + 1 } = 2 \arctan \left( \frac { 4 } { x _ { n } } \right) \quad x _ { 1 } = 2$$
2. find the value of \(x _ { 2 }\) and the value of \(x _ { 6 }\), giving your answers to 3 decimal places.
   

8. \begin{figure}[h] \end{figure} Figure 4 is a graph showing the velocity of a sprinter during a 100 m race.
The sprinter's velocity during the race, \(v \mathrm {~ms} ^ { - 1 }\), is modelled by the equation $$v = 12 - \mathrm { e } ^ { t - 10 } - 12 \mathrm { e } ^ { - 0.75 t } \quad t \geqslant 0$$ where \(t\) seconds is the time after the sprinter begins to run. According to the model,

1. find, using calculus, the sprinter's maximum velocity during the race. Given that the sprinter runs 100 m in \(T\) seconds, such that $$\int _ { 0 } ^ { T } v \mathrm {~d} t = 100$$
2. show that \(T\) is a solution of the equation $$T = \frac { 1 } { 12 } \left( 116 - 16 \mathrm { e } ^ { - 0.75 T } + \mathrm { e } ^ { T - 10 } - \mathrm { e } ^ { - 10 } \right)$$ The iteration formula $$T _ { n + 1 } = \frac { 1 } { 12 } \left( 116 - 16 \mathrm { e } ^ { - 0.75 T _ { n } } + \mathrm { e } ^ { T _ { n } - 10 } - \mathrm { e } ^ { - 10 } \right)$$ is used to find an approximate value for \(T\) Using this iteration formula with \(T _ { 1 } = 10\)
3. find, to 4 decimal places,
   1. the value of \(T _ { 2 }\)
   2. the time taken by the sprinter to run the race, according to the model.

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.

8. \begin{figure}[h] \end{figure} Figure 4 is a graph showing the path of a golf ball after the ball has been hit until it first hits the ground. The vertical height, \(h\) metres, of the ball above the ground has been plotted against the horizontal distance travelled, \(x\) metres, measured from where the ball was hit. The ball travels a horizontal distance of \(d\) metres before it first hits the ground.
The ball is modelled as a particle travelling in a vertical plane above horizontal ground.
The path of the ball is modelled by the equation $$h = 1.5 x - 0.5 x \mathrm { e } ^ { 0.02 x } \quad 0 \leqslant x \leqslant d$$ \section*{Use the model to answer parts (a), (b) and (c).}

1. Find the value of \(d\), giving your answer to 2 decimal places.
   (Solutions relying entirely on calculator technology are not acceptable.)
2. Show that the maximum value of \(h\) occurs when $$x = 50 \ln \left( \frac { 150 } { x + 50 } \right)$$ Using the iteration formula $$x _ { n + 1 } = 50 \ln \left( \frac { 150 } { x _ { n } + 50 } \right) \quad \text { with } x _ { 1 } = 30$$
3. 1. find the value of \(x _ { 2 }\) to 2 decimal places,
   2. find, by repeated iteration, the horizontal distance travelled by the golf ball before it reaches its maximum height. Give your answer to 2 decimal places. \includegraphics[max width=\textwidth, alt={}, center]{5a695b86-1660-4c06-ac96-4cdb07af9a2e-26_2270_56_309_1981}

6. \begin{figure}[h] \end{figure} Figure 3 shows a sketch of curve \(C _ { 1 }\) with equation \(y = 5 \mathrm { e } ^ { x - 1 } + 3\) and curve \(C _ { 2 }\) with equation \(y = 10 - x ^ { 2 }\) The point \(P\) lies on \(C _ { 1 }\) and has \(y\) coordinate 18

1. Find the \(x\) coordinate of \(P\), writing your answer in the form \(\ln k\), where \(k\) is a constant to be found. The curve \(C _ { 1 }\) meets the curve \(C _ { 2 }\) at \(x = \alpha\) and at \(x = \beta\), as shown in Figure 3.
2. Using a suitable interval and a suitable function that should be stated, show that to 3 decimal places \(\alpha = 1.134\) The iterative equation $$x _ { n + 1 } = - \sqrt { 7 - 5 \mathrm { e } ^ { x _ { n } - 1 } }$$ is used to find an approximation to \(\beta\). Using this iterative formula with \(x _ { 1 } = - 3\)
3. find the value of \(x _ { 2 }\) and the value of \(\beta\), giving each answer to 6 decimal places.

9. \begin{figure}[h] \end{figure} Figure 3 shows a sketch of part of the curve with equation \(y = \mathrm { f } ( x )\), where $$f ( x ) = x \left( x ^ { 2 } - 4 \right) e ^ { - \frac { 1 } { 2 } x }$$

1. Find \(f ^ { \prime } ( x )\). The line \(l\) is the normal to the curve at \(O\) and meets the curve again at the point \(P\). The point \(P\) lies in the 3rd quadrant, as shown in Figure 3.
2. Show that the \(x\) coordinate of \(P\) is a solution of the equation $$x = - \frac { 1 } { 2 } \sqrt { 16 + \mathrm { e } ^ { \frac { 1 } { 2 } x } }$$
3. Using the iterative formula $$x _ { n + 1 } = - \frac { 1 } { 2 } \sqrt { 16 + \mathrm { e } ^ { \frac { 1 } { 2 } x _ { n } } } \quad \text { with } x _ { 1 } = - 2$$ find, to 4 decimal places,
   1. the value of \(x _ { 2 }\)
   2. the \(x\) coordinate of \(P\).

5. \begin{figure}[h] \end{figure} The profit made by a company, \(\pounds P\) million, \(t\) years after the company started trading, is modelled by the equation $$P = \frac { 4 t - 1 } { 10 } + \frac { 3 } { 4 } \ln \left[ \frac { t + 1 } { ( 2 t + 1 ) ^ { 2 } } \right]$$ The graph of \(P\) against \(t\) is shown in Figure 2. According to the model,

1. show that exactly one year after it started trading, the company had made a loss of approximately £ 830000 A manager of the company wants to know the value of \(t\) for which \(P = 0\)
2. Show that this value of \(t\) occurs in the interval [6,7]
3. Show that the equation \(P = 0\) can be expressed in the form $$t = \frac { 1 } { 4 } + \frac { 15 } { 8 } \ln \left[ \frac { ( 2 t + 1 ) ^ { 2 } } { t + 1 } \right]$$
4. Using the iteration formula $$t _ { n + 1 } = \frac { 1 } { 4 } + \frac { 15 } { 8 } \ln \left[ \frac { \left( 2 t _ { n } + 1 \right) ^ { 2 } } { t _ { n } + 1 } \right] \text { with } t _ { 1 } = 6$$ find the value of \(t _ { 2 }\) and the value of \(t _ { 6 }\), giving your answers to 3 decimal places.
5. Hence find, according to the model, how many months it takes in total, from when the company started trading, for it to make a profit.
   (2)

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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7. \begin{figure}[h] \end{figure} Figure 3 shows a sketch of part of the curve with equation \(y = \mathrm { f } ( x )\), where $$f ( x ) = 2 x ( 1 + x ) \ln x , \quad x > 0$$ The curve has a minimum turning point at \(A\).

1. Find f'(x)
2. Hence show that the \(x\) coordinate of \(A\) is the solution of the equation $$x = \mathrm { e } ^ { - \frac { 1 + x } { 1 + 2 x } }$$
3. Use the iteration formula $$x _ { n + 1 } = \mathrm { e } ^ { - \frac { 1 + x _ { n } } { 1 + 2 x _ { n } } } , \quad x _ { 0 } = 0.46$$ to find the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\) to 4 decimal places.
4. Use your answer to part (c) to estimate the coordinates of \(A\) to 2 decimal places.

10. \begin{figure}[h] \end{figure} Figure 3 shows a sketch of part of the curve \(C\) with equation $$y = \frac { x ^ { 2 } \ln x } { 3 } - 2 x + 4 , \quad x > 0$$ Point \(A\) is the minimum turning point on the curve.

1. Show, by using calculus, that the \(x\) coordinate of point \(A\) is a solution of $$x = \frac { 6 } { 1 + \ln \left( x ^ { 2 } \right) }$$
2. Starting with \(x _ { 0 } = 2.27\), use the iteration $$x _ { n + 1 } = \frac { 6 } { 1 + \ln \left( x _ { n } ^ { 2 } \right) }$$ 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 point \(A\) to one decimal place.

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.

11.

1. Given that \(0 \leqslant \mathrm { f } ( x ) \leqslant \pi\), sketch the graph of \(y = \mathrm { f } ( x )\) where $$\mathrm { f } ( x ) = \arccos ( x - 1 ) , \quad 0 \leqslant x \leqslant 2$$ The equation \(\arccos ( x - 1 ) - \tan x = 0\) has a single root \(\alpha\).
2. Show that \(0.9 < \alpha < 1.1\) The iteration formula $$x _ { n + 1 } = \arctan \left( \arccos \left( x _ { n } - 1 \right) \right)$$ can be used to find an approximation for \(\alpha\).
3. Taking \(x _ { 0 } = 1.1\) find, to 3 decimal places, the values of \(x _ { 1 }\) and \(x _ { 2 }\)

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.

11. $$y = \left( 2 x ^ { 2 } - 3 \right) \tan \left( \frac { 1 } { 2 } x \right) , \quad 0 < x < \pi$$

1. Find the exact value of \(x\) when \(y = 0\) Given that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\) when \(x = \alpha\),
2. show that $$2 \alpha ^ { 2 } - 3 + 4 \alpha \sin \alpha = 0$$ The iterative formula $$x _ { n + 1 } = \frac { 3 } { \left( 2 x _ { n } + 4 \sin x _ { n } \right) }$$ can be used to find an approximation for \(\alpha\).
3. Taking \(x _ { 1 } = 0.7\), find the values of \(x _ { 2 }\) and \(x _ { 3 }\), giving each answer to 4 decimal places.
4. By choosing a suitable interval, show that \(\alpha = 0.7283\) to 4 decimal places.
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3. $$f ( x ) = 2 ^ { x - 1 } - 4 + 1.5 x \quad x \in \mathbb { R }$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) can be written as $$x = \frac { 1 } { 3 } \left( 8 - 2 ^ { x } \right)$$ The equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\), where \(\alpha = 1.6\) to one decimal place.
2. Starting with \(x _ { 0 } = 1.6\), use the iteration formula $$x _ { n + 1 } = \frac { 1 } { 3 } \left( 8 - 2 ^ { x _ { n } } \right)$$ to calculate the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\), giving your answers to 3 decimal places.
3. By choosing a suitable interval, prove that \(\alpha = 1.633\) 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 $$2 ( x + 2 ) ^ { 3 } + x - 18 = 0$$

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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.

4. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the curve with equation \(y = 8 x - x \mathrm { e } ^ { 3 x } , x \geqslant 0\) The curve meets the \(x\)-axis at the origin and cuts the \(x\)-axis at the point \(A\).

1. Find the exact \(x\) coordinate of \(A\), giving your answer in its simplest form. The curve has a maximum turning point at the point \(M\).
2. Show, by using calculus, that the \(x\) coordinate of \(M\) is a solution of $$x = \frac { 1 } { 3 } \ln \left( \frac { 8 } { 1 + 3 x } \right)$$
3. Use the iterative formula $$x _ { n + 1 } = \frac { 1 } { 3 } \ln \left( \frac { 8 } { 1 + 3 x _ { n } } \right)$$ with \(x _ { 0 } = 0.4\) to calculate the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\), giving your answers to 3 decimal places.