1.09b Sign change methods: understand failure cases

6. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the curve with equation $$f ( x ) = 4 \cos 2 x - 2 x + 1 \quad x > 0$$ and where \(x\) is measured in radians.
The curve crosses the \(x\)-axis at the point \(A\), as shown in figure 1 .
Given that \(x\)-coordinate of \(A\) is \(\alpha\) a. show that \(\alpha\) lies between 0.7 and 0.8 Given that \(x\)-coordinates of \(B\) and \(C\) are \(\beta\) and \(\gamma\) respectively and they are two smallest values of \(x\) at which local maxima occur
b. find, using calculus, the value of \(\beta\) and the value of \(\gamma\), giving your answers to 3 significant figures.
c. taking \(x _ { 0 } = 0.7\) or 0.8 as a first approximation to \(\alpha\), apply the Newton-Raphson method once to \(\mathrm { f } ( x )\) to obtain a second approximation to \(\alpha\). Show, your method and give your answer to 2 significant figures.

2. \begin{figure}[h] \end{figure} Figure 1 shows a plot of part of the curve with equation \(y = \cos x\) where \(x\) is measured in radians. Diagram 1, on the opposite page, is a copy of Figure 1.

1. Use Diagram 1 to show why the equation $$\cos x - 2 x - \frac { 1 } { 2 } = 0$$ has only one real root, giving a reason for your answer. Given that the root of the equation is \(\alpha\), and that \(\alpha\) is small,
2. use the small angle approximation for \(\cos x\) to estimate the value of \(\alpha\) to 3 decimal places.
   \includegraphics[max width=\textwidth, alt={}]{91a2f26a-add2-4b58-997d-2ae229548217-05_664_1452_246_333}
   \section*{Diagram 1}

9 This question is about the equation \(\mathrm { f } ( x ) = 0\), where \(\mathrm { f } ( x ) = x ^ { 4 } - x - \frac { 1 } { 3 x - 2 }\).
Fig. 9.1 shows the curve \(y = f ( x )\).
Fig. 9.1 \includegraphics[max width=\textwidth, alt={}, center]{60e1e785-c34b-48ef-a63f-13a25fee186e-06_940_929_518_239}

1. Show, by calculation, that the equation \(\mathrm { f } ( x ) = 0\) has a root between \(x = 1\) and \(x = 2\).
2. Fig. 9.2 shows part of a spreadsheet being used to find a root of the equation. \begin{table}[h]
   \captionsetup{labelformat=empty} \caption{Fig. 9.2}

   	A	B
   1	\(x\)	\(f ( x )\)
   2	1.5	3.1625
   3	1.25	0.619977679
   4	1.125	- 0.250466087
   5

   \end{table} Write down a suitable number to use as the next value of \(x\) in the spreadsheet.
3. Determine a root of the equation \(\mathrm { f } ( x ) = 0\). Give your answer correct to \(\mathbf { 1 }\) decimal place.
4. Fig. 9.3 shows a similar spreadsheet being used to search for another root of \(\mathrm { f } ( x ) = 0\). \begin{table}[h]
   \captionsetup{labelformat=empty} \caption{Fig. 9.3}

   	A	B
   1	x	f(x)
   2	0	0.5
   3	1	-1
   4	0.5	1.5625
   5	0.75	-4.4336
   6	0.6	4.5296
   7	0.7	-10.4599
   8	0.65	19.5285
   9	0.675	-40.4674
   10	0.6625	79.5301
   11	0.66875	-160.4687
   10

   \end{table}
   1. Explain why it looks from rows 2 and 3 of the spreadsheet as if there is a root between 0 and 1.
   2. Explain why this process will not find a root between 0 and 1 .

10 The function \(\mathrm { f } ( x )\) is defined by \(\mathrm { f } ( x ) = x ^ { 4 } + x ^ { 3 } - 2 x ^ { 2 } - 4 x - 2\).

1. Show that \(x = - 1\) is a root of \(\mathrm { f } ( x ) = 0\).
2. Show that another root of \(\mathrm { f } ( x ) = 0\) lies between \(x = 1\) and \(x = 2\).
3. Show that \(\mathrm { f } ( x ) = ( x + 1 ) \mathrm { g } ( x )\), where \(\mathrm { g } ( x ) = x ^ { 3 } + a x + b\) and \(a\) and \(b\) are integers to be determined.
4. Without further calculation, explain why \(\mathrm { g } ( x ) = 0\) has a root between \(x = 1\) and \(x = 2\).
5. Use the Newton-Raphson formula to show that an iteration formula for finding roots of \(\mathrm { g } ( x ) = 0\) may be written $$x _ { n + 1 } = \frac { 2 x _ { n } ^ { 3 } + 2 } { 3 x _ { n } ^ { 2 } - 2 }$$ Determine the root of \(\mathrm { g } ( x ) = 0\) which lies between \(x = 1\) and \(x = 2\) correct to 4 significant figures.

8 The diagram shows the curve \(y = \cos ^ { - 1 } x\) for \(- 1 \leqslant x \leqslant 1\). \includegraphics[max width=\textwidth, alt={}, center]{6890a681-2b7f-4853-a5f0-f88b7b435367-4_492_698_1640_671}

1. Write down the exact coordinates of the points \(A\) and \(B\).
2. The equation \(\cos ^ { - 1 } x = 3 x + 1\) has only one root. Given that the root of this equation is \(\alpha\), show that \(0.1 \leqslant \alpha \leqslant 0.2\).
3. Use the iteration \(x _ { n + 1 } = \frac { 1 } { 3 } \left( \cos ^ { - 1 } x _ { n } - 1 \right)\) with \(x _ { 1 } = 0.1\) to find the values of \(x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving your answers to three decimal places.

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.

8. A curve has the equation \(y = \frac { \mathrm { e } ^ { 2 } } { x } + \mathrm { e } ^ { x } , \quad x \neq 0\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
   [0pt]
2. Show that the curve has a stationary point in the interval [1.3,1.4]. The point \(A\) on the curve has \(x\)-coordinate 2 .
3. Show that the tangent to the curve at \(A\) passes through the origin. The tangent to the curve at \(A\) intersects the curve again at the point \(B\).
   The \(x\)-coordinate of \(B\) is to be estimated using the iterative formula $$x _ { n + 1 } = - \frac { 2 } { 3 } \sqrt { 3 + 3 x _ { n } \mathrm { e } ^ { x _ { n } - 2 } }$$ with \(x _ { 0 } = - 1\).
4. Find \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\) to 7 significant figures and hence state the \(x\)-coordinate of \(B\) to 5 significant figures.

8. $$f ( x ) = 2 x + \sin x - 3 \cos x$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root in the interval [0.7, 0.8].
2. Find an equation for the tangent to the curve \(y = \mathrm { f } ( x )\) at the point where it crosses the \(y\)-axis.
3. Find the values of the constants \(a , b\) and \(c\), where \(b > 0\) and \(0 < c < \frac { \pi } { 2 }\), such that $$f ^ { \prime } ( x ) = a + b \cos ( x - c )$$
4. Hence find the \(x\)-coordinates of the stationary points of the curve \(y = \mathrm { f } ( x )\) in the interval \(0 \leq x \leq 2 \pi\), giving your answers to 2 decimal places.

7. The functions \(f\) and \(g\) are defined by $$\begin{aligned} & \mathrm { f } : x \rightarrow | 2 x - 5 | , \quad x \in \mathbb { R } , \\ & \mathrm {~g} : x \rightarrow \ln ( x + 3 ) , \quad x \in \mathbb { R } , \quad x > - 3 \end{aligned}$$

1. State the range of f .
2. Evaluate fg(-2).
3. Solve the equation $$\operatorname { fg } ( x ) = 3$$ giving your answers in exact form.
4. Show that the equation $$\mathrm { f } ( x ) = \mathrm { g } ( x )$$ has a root, \(\alpha\), in the interval [3,4].
5. Use the iteration formula $$x _ { n + 1 } = \frac { 1 } { 2 } \left[ 5 + \ln \left( x _ { n } + 3 \right) \right]$$ with \(x _ { 0 } = 3\), to find \(x _ { 1 } , x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving your answers to 4 significant figures.
6. Show that your answer for \(x _ { 4 }\) is the value of \(\alpha\) correct to 4 significant figures. \end{document}

7 The equation $$24 x ^ { 3 } + 36 x ^ { 2 } + 18 x - 5 = 0$$ has one real root, \(\alpha\).

1. Show that \(\alpha\) lies in the interval \(0.1 < x < 0.2\).
2. Starting from the interval \(0.1 < x < 0.2\), use interval bisection twice to obtain an interval of width 0.025 within which \(\alpha\) must lie.
3. Taking \(x _ { 1 } = 0.2\) as a first approximation to \(\alpha\), use the Newton-Raphson method to find a second approximation, \(x _ { 2 }\), to \(\alpha\). Give your answer to four decimal places.
   (4 marks)

The curve \(C\) has equation \(y = f(x)\), where $$f(x) = 3 \ln x + \frac{1}{x}, \quad x > 0.$$ The point \(P\) is a stationary point on \(C\).

1. Calculate the \(x\)-coordinate of \(P\). [4]
2. Show that the \(y\)-coordinate of \(P\) may be expressed in the form \(k - k \ln k\), where \(k\) is a constant to be found. [2]

The point \(Q\) on \(C\) has \(x\)-coordinate \(1\).

1. Find an equation for the normal to \(C\) at \(Q\). [4]

The normal to \(C\) at \(Q\) meets \(C\) again at the point \(R\).

1. Show that the \(x\)-coordinate of \(R\)
   1. satisfies the equation \(6 \ln x + x + \frac{2}{x} - 3 = 0\),
   2. lies between \(0.13\) and \(0.14\). [4]

$$f(x) = 2x^{-\frac{2}{3}} + \frac{1}{2}x - \frac{1}{3x - 5} - \frac{5}{2} \quad x \neq \frac{5}{3}$$ The table below shows values of \(f(x)\) for some values of \(x\), with values of \(f(x)\) given to 4 decimal places where appropriate.

1. Complete the table giving the values to 4 decimal places. [2]

The equation \(f(x) = 0\) has exactly one positive root, \(\alpha\). Using the values in the completed table and explaining your reasoning,

1. determine an interval of width one that contains \(\alpha\). [2]
2. Hence use interval bisection twice to obtain an interval of width 0.25 that contains \(\alpha\). [3]

Given also that the equation \(f(x) = 0\) has a negative root, \(\beta\), in the interval \([-1, -0.5]\)

1. use linear interpolation once on this interval to find an approximation for \(\beta\). Give your answer to 3 significant figures. [3]

$$f(x) = \frac{1}{2}x^4 - x^3 + x - 3$$

1. Show that the equation \(f(x) = 0\) has a root \(\alpha\) between \(x = 2\) and \(x = 2.5\) [2]
2. Starting with the interval \([2, 2.5]\) use interval bisection twice to find an interval of width \(0.125\) which contains \(\alpha\). [3]

The equation \(f(x) = 0\) has a root \(\beta\) in the interval \([-2, -1]\).

1. Taking \(-1.5\) as a first approximation to \(\beta\), apply the Newton-Raphson process once to \(f(x)\) to obtain a second approximation to \(\beta\). Give your answer to 2 decimal places. [5]

\(\text{f}(x) = x^3 + x^2 - 4x - 1\). The equation f(x) = 0 has only one positive root, \(\alpha\).

1. Show that f(x) = 0 can be rearranged as $$x = \sqrt{\frac{4x+1}{x+1}}, \quad x \neq -1.$$ [2]

The iterative formula \(x_{n+1} = \sqrt{\frac{4x_n+1}{x_n+1}}\) is used to find an approximation to \(\alpha\).

1. Taking \(x_1 = 1\), find, to 2 decimal places, the values of \(x_2\), \(x_3\) and \(x_4\). [3]
2. By choosing values of \(x\) in a suitable interval, prove that \(\alpha = 1.70\), correct to 2 decimal places. [3]
3. Write down a value of \(x_1\) for which the iteration formula \(x_{n+1} = \sqrt{\frac{4x_n+1}{x_n+1}}\) does not produce a valid value for \(x_2\). Justify your answer. [2]

In this question you must show detailed reasoning. \includegraphics{figure_5} The diagram shows the curve with equation \(y = \frac{2x - 3}{4x^2 + 1}\). The tangent to the curve at the point \(P\) has gradient 2.

1. Show that the \(x\)-coordinate of \(P\) satisfies the equation $$4x^3 + 3x - 3 = 0.$$ [5]
2. Show by calculation that the \(x\)-coordinate of \(P\) lies between 0.5 and 1. [2]
3. Show that the iteration $$x_{n+1} = \frac{3 - 4x_n^3}{3}$$ cannot converge to the \(x\)-coordinate of \(P\) whatever starting value is used. [2]
4. Use the Newton-Raphson method, with initial value 0.5, to determine the coordinates of \(P\) correct to 5 decimal places. [5]