Explain Sign Change Method Failure

10
\includegraphics[max width=\textwidth, alt={}, center]{9473b8f7-616a-485e-963b-696c6640ae6b-07_805_775_251_242} The diagram shows part of the curve \(\mathrm { f } ( x ) = \frac { \mathrm { e } ^ { x } } { 4 x ^ { 2 } - 1 } + 2\). The equation \(\mathrm { f } ( x ) = 0\) has a positive root \(\alpha\) close to \(x = 0.3\).

1. Explain why using the sign change method with \(x = 0\) and \(x = 1\) will fail to locate \(\alpha\).
2. Show that the equation \(\mathrm { f } ( x ) = 0\) can be written as \(x = \frac { 1 } { 4 } \sqrt { \left( 4 - 2 \mathrm { e } ^ { x } \right) }\).
3. Use the iterative formula \(x _ { n + 1 } = \frac { 1 } { 4 } \sqrt { \left( 4 - 2 \mathrm { e } ^ { x _ { n } } \right) }\) with a starting value of \(x _ { 1 } = 0.3\) to find the value of \(\alpha\) correct to \(\mathbf { 4 }\) significant figures, showing the result of each iteration.
4. An alternative iterative formula is \(x _ { n + 1 } = \mathrm { F } \left( x _ { n } \right)\), where \(\mathrm { F } \left( x _ { n } \right) = \ln \left( 2 - 8 x _ { n } ^ { 2 } \right)\). By considering \(\mathrm { F } ^ { \prime } ( 0.3 )\) explain why this iterative formula will not find \(\alpha\).

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.

2 Fig. 2.1 shows the graph of \(y = x ^ { 2 } \mathrm { e } ^ { 2 x } - 5 x ^ { 2 } + 0.5\). \begin{figure}[h] \end{figure} There are three roots of the equation \(x ^ { 2 } \mathrm { e } ^ { 2 x } - 5 x ^ { 2 } + 0.5 = 0\). The roots are \(\alpha , \beta\) and \(\gamma\), where \(\alpha < \beta < \gamma\).

1. Explain why it is not possible to use the method of false position with \(x _ { 0 } = 0\) and \(x _ { 1 } = 1\) to find \(\beta\) and \(\gamma\). The graph of the function indicates that the root \(\gamma\) lies in the interval [0.6, 0.8]. Fig. 2.2 shows some spreadsheet output using the method of false position using these values as starting points. \begin{table}[h]

   	A	B	C	D	E	F
   1	a	f(a)	b	f(b)	approx
   2	0.6	-0.10476	0.8	0.469941	0.636457	-0.07876
   3	0.636457	-0.07876	0.8	0.469941	0.659931	-0.04748
   4	0.659931	-0.04748	0.8	0.469941	0.672783	-0.0249
   5	0.672783	-0.0249	0.8	0.469941	0.679184	-0.01211
   6	0.679184	-0.01211	0.8	0.469941	0.682218	-0.00567
   7	0.682218	-0.00567	0.8	0.469941	0.683623	-0.00261
   8	0.683623	-0.00261	0.8	0.469941	0.684266	-0.00119
   9	0.684266	-0.00119	0.8	0.469941	0.684559	-0.00054
   10	0.684559	-0.00054	0.8	0.469941	0.684692	-0.00025
   11	0.684692	-0.00025	0.8	0.469941	0.684753	-0.00011
   12	0.684753	-0.00011	0.8	0.469941	0.68478	\(- 5.1 \mathrm { E } - 05\)

   \captionsetup{labelformat=empty} \caption{Fig. 2.2}
   \end{table}
2. Without doing any further calculation, write down the smallest possible interval which is certain to contain \(\gamma\).
3. State what is being calculated in column F. The formula in cell A3 is \(\quad = \operatorname { IF } ( \mathrm { F } 2 < 0 , \mathrm { E } 2 , \mathrm {~A} 2 )\).
4. Explain the purpose of this formula in the application of the method of false position. The method of false position uses the same formula for obtaining new approximations as the secant method.
5. Explain how the method of false position differs from the secant method.
6. Give one advantage and one disadvantage of using the method of false position instead of the secant method.

2 A student is searching for a solution to the equation \(\mathrm { f } ( x ) = 0\) He correctly evaluates $$f ( - 1 ) = - 1 \text { and } f ( 1 ) = 1$$ and concludes that there must be a root between - 1 and 1 due to the change of sign.
Select the function \(\mathrm { f } ( x )\) for which the conclusion is incorrect.
Circle your answer. $$\mathrm { f } ( x ) = \frac { 1 } { x } \quad \mathrm { f } ( x ) = x \quad \mathrm { f } ( x ) = x ^ { 3 } \quad \mathrm { f } ( x ) = \frac { 2 x + 1 } { x + 2 }$$