Sign Change & Interval Methods

$$f(x) = 1 - e^x + 3 \sin 2x$$ The equation \(f(x) = 0\) has a root \(\alpha\) in the interval \(1.0 < x < 1.4\). Starting with the interval \((1.0, 1.4)\), use interval bisection three times to find the value of \(\alpha\) to one decimal place. [3]

$$f(x) = 2^x + x - 4.$$ The equation \(f(x) = 0\) has a root \(\alpha\) in the interval \([1, 2]\). Use linear interpolation on the values at the end points of this interval to find an approximation to \(\alpha\). [2]

2 By considering a change of sign, show that the equation \(\mathrm { e } ^ { x } - 5 x ^ { 3 } = 0\) has a root between 0 and 1 .

2 The diagram shows part of the graph of \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x )\) is a cubic polynomial in \(x\). \includegraphics[max width=\textwidth, alt={}, center]{7298e7b9-ad52-480c-bc2b-8289aeab9ebb-04_437_620_909_274} Explain why one of the roots of the equation \(\mathrm { f } ( x ) = 0\) cannot be found by the sign change method.

8 The graph of \(\mathrm { y } = 0.2 \cosh \mathrm { x } - 0.4 \mathrm { x }\) for values of \(x\) from 0 to 3.32 is shown on the graph below. \includegraphics[max width=\textwidth, alt={}, center]{4023e87c-34b1-4abd-9acc-ede5e4d68c7f-08_988_1561_312_244} The equation \(0.2 \cosh x - 0.4 x = 0\) has two roots, \(\alpha\) and \(\beta\) where \(\alpha < \beta\), in the interval \(0 < x < 3\). The secant method with \(x _ { 0 } = 1\) and \(x _ { 1 } = 2\) is to be used to find \(\beta\).

1. On the copy of the graph in the Printed Answer Booklet, show how the secant method works with these two values of \(x\) to obtain an improved approximation to \(\beta\). The spreadsheet output in the table below shows the result of applying the secant method with \(x _ { 0 } = 1\) and \(x _ { 1 } = 2\).

   	I	J	K	L	M
   2	\(r\)	\(\mathrm { x } _ { \mathrm { r } }\)	f(x)	\(\mathrm { X } _ { \mathrm { r } + 1 }\)	\(\mathrm { f } \left( \mathrm { x } _ { \mathrm { r } + 1 } \right)\)
   3	0	1	-0.0914	2	-0.0476
   4	1	2	-0.0476	3.08529	0.95784
   5	2	3.08529	0.95784	2.05134	-0.0298
   6	3	2.05134	-0.0298	2.08259	-0.0181
   7	4	2.08259	-0.0181	2.13042	0.00155
   8	5	2.13042	0.00155	2.12664	\(- 7 \mathrm { E } - 05\)

2. Write down a suitable cell formula for cell J4.
3. Write down a suitable cell formula for cell L4.
4. Write down the most accurate approximation to \(\beta\) which is displayed in the table.
5. Determine whether your answer to part (d) is correct to 5 decimal places. You should not calculate any more iterates.
6. It is decided to use the secant method with starting values \(x _ { 0 } = 1\) and \(\mathrm { x } _ { 1 } = \mathrm { a }\), where \(a > 1\), to find \(\alpha\). State a suitable value for \(a\).

2. (a) Show that \(\mathrm { f } ( x ) = x ^ { 4 } + x - 1\) has a real root \(\alpha\) in the interval [0.5, 1.0].
[0pt] (b) Starting with the interval [0.5, 1.0], use interval bisection twice to find an interval of width 0.125 which contains \(\alpha\).
(c) Taking 0.75 as a first approximation, apply the Newton Raphson process twice to \(\mathrm { f } ( x )\) to obtain an approximate value of \(\alpha\). Give your answer to 3 decimal places.

2 The following spreadsheet printout shows the bisection method being applied to the equation \(\mathrm { f } ( x ) = 0\), where \(\mathrm { f } ( x ) = \mathrm { e } ^ { x } - x ^ { 2 } - 2\). Some values of \(\mathrm { f } ( x )\) are shown in columns B and D.

1. The formula in cell A 3 is \(= \mathrm { IF } ( \mathrm { F } 2 > 0\), A2, E2). State the purpose of this formula.
2. The formula in cell C 3 is \(= \mathrm { IF } ( \mathrm { F } 2 > 0 , \ldots , \ldots )\). What are the missing cell references?
3. In which row is the magnitude of the maximum possible error (mpe) less than \(5 \times 10 ^ { - 7 }\) for the first time?

9. The diagram shows the curve with equation \(y = 2 x - 3 \ln ( 2 x + 5 )\) and the normal to the curve at the point \(P ( - 2 , - 4 )\).

1. Find an equation for the normal to the curve at \(P\). The normal to the curve at \(P\) intersects the curve again at the point \(Q\) with \(x\)-coordinate \(q\).
2. Show that \(1 < q < 2\).
3. Show that \(q\) is a solution of the equation $$x = \frac { 12 } { 7 } \ln ( 2 x + 5 ) - 2 .$$
4. Use an iterative process based on the equation above with a starting value of 1.5 to find the value of \(q\) to 3 significant figures and justify the accuracy of your answer.