Sketch Graphs and Count Roots

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1. 
   \includegraphics[max width=\textwidth, alt={}, center]{4ce3208e-8ceb-4848-a9c7-fcda166319f4-05_753_944_278_630} The diagram shows the graph of \(y = 3 - \mathrm { e } ^ { - \frac { 1 } { 2 } x }\).
   On the diagram, sketch the graph of \(y = | 5 x - 4 |\), and show that the equation \(3 - e ^ { - \frac { 1 } { 2 } x } = | 5 x - 4 |\) has exactly two real roots. It is given that the two roots of \(3 - \mathrm { e } ^ { - \frac { 1 } { 2 } x } = | 5 x - 4 |\) are denoted by \(\alpha\) and \(\beta\), where \(\alpha < \beta\).
2. Show by calculation that \(\alpha\) lies between 0.36 and 0.37 .
3. Use the iterative formula \(x _ { n + 1 } = \frac { 1 } { 5 } \left( 7 - \mathrm { e } ^ { - \frac { 1 } { 2 } x _ { n } } \right)\) to find \(\beta\) correct to 4 significant figures. Give the result of each iteration to 6 significant figures.

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1. 
   \includegraphics[max width=\textwidth, alt={}, center]{a1ea242a-c7f4-46b0-b4b8-bd13b3880557-05_753_944_278_630} The diagram shows the graph of \(y = 3 - \mathrm { e } ^ { - \frac { 1 } { 2 } x }\).
   On the diagram, sketch the graph of \(y = | 5 x - 4 |\), and show that the equation \(3 - e ^ { - \frac { 1 } { 2 } x } = | 5 x - 4 |\) has exactly two real roots. It is given that the two roots of \(3 - \mathrm { e } ^ { - \frac { 1 } { 2 } x } = | 5 x - 4 |\) are denoted by \(\alpha\) and \(\beta\), where \(\alpha < \beta\).
2. Show by calculation that \(\alpha\) lies between 0.36 and 0.37 .
3. Use the iterative formula \(x _ { n + 1 } = \frac { 1 } { 5 } \left( 7 - \mathrm { e } ^ { - \frac { 1 } { 2 } x _ { n } } \right)\) to find \(\beta\) correct to 4 significant figures. Give the result of each iteration to 6 significant figures.

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.

1. (a) On the same diagram, sketch and clearly label the graphs with equations

$$y = \mathrm { e } ^ { x } \quad \text { and } \quad y = 10 - x$$ Show on your sketch the coordinates of each point at which the graphs cut the axes.
(b) Explain why the equation \(\mathrm { e } ^ { x } - 10 + x = 0\) has only one solution.
(c) Show that the solution of the equation $$\mathrm { e } ^ { x } - 10 + x = 0$$ lies between \(x = 2\) and \(x = 3\)
(d) Use the iterative formula $$x _ { n + 1 } = \ln \left( 10 - x _ { n } \right) , \quad x _ { 1 } = 2$$ to calculate the values of \(x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\).
Give your answers to 4 decimal places.

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}

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\includegraphics[max width=\textwidth, alt={}, center]{33a2b09d-0df9-48d6-9ee9-e0a1ec345f41-3_524_720_246_676} The diagram shows the curve \(y = x ^ { 4 } - 8 x\).

1. By sketching a second curve on the copy of the diagram, show that the equation $$x ^ { 4 } + x ^ { 2 } - 8 x - 9 = 0$$ has two real roots. State the equation of the second curve.
2. The larger root of the equation \(x ^ { 4 } + x ^ { 2 } - 8 x - 9 = 0\) is denoted by \(\alpha\).
   (a) Show by calculation that \(2.1 < \alpha < 2.2\).
   (b) Use an iterative process based on the equation $$x = \sqrt [ 4 ] { 9 + 8 x - x ^ { 2 } } ,$$ with a suitable starting value, to find \(\alpha\) correct to 3 decimal places. Give the result of each step of the iterative process.

4. (a) Sketch, on the same set of axes, the graphs of $$y = 2 - \mathrm { e } ^ { - x } \text { and } y = \sqrt { } x$$ [It is not necessary to find the coordinates of any points of intersection with the axes.]
Given that \(\mathrm { f } ( x ) = \mathrm { e } ^ { - x } + \sqrt { } x - 2 , x \geq 0\),
(b) explain how your graphs show that the equation \(\mathrm { f } ( x ) = 0\) has only one solution,
(c) show that the solution of \(\mathrm { f } ( x ) = 0\) lies between \(x = 3\) and \(x = 4\). The iterative formula \(x _ { n + 1 } = \left( 2 - \mathrm { e } ^ { - x _ { n } } \right) ^ { 2 }\) is used to solve the equation \(\mathrm { f } ( x ) = 0\).
(d) Taking \(x _ { 0 } = 4\), write down the values of \(x _ { 1 } , x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), and hence find an approximation to the solution of \(\mathrm { f } ( x ) = 0\), giving your answer to 3 decimal places.

5. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of the curve with equation \(y = \mathrm { e } ^ { - x } - 1\).

1. Copy Fig. 1 and on the same axes sketch the graph of \(y = \frac { 1 } { 2 } | x - 1 |\). Show the coordinates of the points where the graph meets the axes. The \(x\)-coordinate of the point of intersection of the graph is \(\alpha\).
2. Show that \(x = \alpha\) is a root of the equation \(x + 2 \mathrm { e } ^ { - x } - 3 = 0\).
3. Show that \(- 1 < \alpha < 0\). The iterative formula \(x _ { \mathrm { n } + 1 } = - \ln \left[ \frac { 1 } { 2 } \left( 3 - x _ { n } \right) \right]\) is used to solve the equation \(x + 2 \mathrm { e } ^ { - x } - 3 = 0\).
4. Starting with \(x _ { 0 } = - 1\), find the values of \(x _ { 1 }\) and \(x _ { 2 }\).
5. Show that, to 2 decimal places, \(\alpha = - 0.58\).

2 [Figure 1, printed on the insert, is provided for use in this question.]

1. 1. Sketch the graph of \(y = \sin ^ { - 1 } x\), where \(y\) is in radians. State the coordinates of the end points of the graph.
   2. By drawing a suitable straight line on your sketch, show that the equation $$\sin ^ { - 1 } x = \frac { 1 } { 4 } x + 1$$ has only one solution.
2. The root of the equation \(\sin ^ { - 1 } x = \frac { 1 } { 4 } x + 1\) is \(\alpha\). Show that \(0.5 < \alpha < 1\).
3. The equation \(\sin ^ { - 1 } x = \frac { 1 } { 4 } x + 1\) can be rewritten as \(x = \sin \left( \frac { 1 } { 4 } x + 1 \right)\).
   1. Use the iteration \(x _ { n + 1 } = \sin \left( \frac { 1 } { 4 } x _ { n } + 1 \right)\) with \(x _ { 1 } = 0.5\) to find the values of \(x _ { 2 }\) and \(x _ { 3 }\), giving your answers to three decimal places.
   2. The sketch on Figure 1 shows parts of the graphs of \(y = \sin \left( \frac { 1 } { 4 } x + 1 \right)\) 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.