1.09a Sign change methods: locate roots

5 Fig. 5 shows part of the curve \(y = \operatorname { cosec } x\) together with the \(x\) - and \(y\)-axes. \begin{figure}[h] \end{figure}

1. For the section of the curve which is shown in Fig. 5, write down
   1. the equations of the two vertical asymptotes,
   2. the coordinates of the minimum point.
2. Show that the equation \(x = \operatorname { cosec } x\) has a root which lies between \(x = 1\) and \(x = 2\).
3. Use the iteration \(\mathrm { x } _ { \mathrm { n } + 1 } = \operatorname { cosec } \left( \mathrm { x } _ { \mathrm { n } } \right)\), with \(x _ { 0 } = 1\), to find
   1. the values of \(x _ { 1 }\) and \(x _ { 2 }\), correct to 5 decimal places,
   2. this root of the equation, correct to 3 decimal places.
4. There is another root of \(x = \operatorname { cosec } x\) which lies between \(x = 2\) and \(x = 3\). Determine whether the iteration \(\mathrm { x } _ { \mathrm { n } + 1 } = \operatorname { cosec } \left( \mathrm { x } _ { \mathrm { n } } \right)\) with \(x _ { 0 } = 2.5\) converges to this root.
5. Sketch the staircase or cobweb diagram for the iteration, starting with \(x _ { 0 } = 2.5\), on the diagram in the Printed Answer Booklet.

2 A curve is defined by the equation \(y = \left( x ^ { 2 } - 4 \right) \ln ( x + 2 )\) for \(x \geqslant 3\).
The curve intersects the line \(y = 15\) at a single point, where \(x = \alpha\).

1. Show that \(\alpha\) lies between 3.5 and 3.6.
2. Show that the equation \(\left( x ^ { 2 } - 4 \right) \ln ( x + 2 ) = 15\) can be arranged into the form $$x = \pm \sqrt { 4 + \frac { 15 } { \ln ( x + 2 ) } }$$ (2 marks)
3. Use the iteration $$x _ { n + 1 } = \sqrt { 4 + \frac { 15 } { \ln \left( x _ { n } + 2 \right) } }$$ with \(x _ { 1 } = 3.5\) to find the values of \(x _ { 2 }\) and \(x _ { 3 }\), giving your answers to three decimal places.
   (2 marks)

1

1. Use Simpson's rule with 7 ordinates (6 strips) to find an estimate for \(\int _ { 0 } ^ { 3 } 4 ^ { x } \mathrm {~d} x\).
2. A curve is defined by the equation \(y = 4 ^ { x }\). The curve intersects the line \(y = 8 - 2 x\) at a single point where \(x = \alpha\).
   1. Show that \(\alpha\) lies between 1.2 and 1.3.
   2. The equation \(4 ^ { x } = 8 - 2 x\) can be rearranged into the form \(x = \frac { \ln ( 8 - 2 x ) } { \ln 4 }\). Use the iterative formula \(x _ { n + 1 } = \frac { \ln \left( 8 - 2 x _ { n } \right) } { \ln 4 }\) with \(x _ { 1 } = 1.2\) to find the values of \(x _ { 2 }\) and \(x _ { 3 }\), giving your answers to three decimal places.
      (2 marks)

1

1. Show that the equation \(x ^ { 3 } - 6 x + 1 = 0\) has a root \(\alpha\), where \(2 < \alpha < 3\).
2. Show that the equation \(x ^ { 3 } - 6 x + 1 = 0\) can be rearranged into the form $$x ^ { 2 } = 6 - \frac { 1 } { x }$$ (1 mark)
3. Use the recurrence relation \(x _ { n + 1 } = \sqrt { 6 - \frac { 1 } { x _ { n } } }\), with \(x _ { 1 } = 2.5\), to find the value of \(x _ { 3 }\), giving your answer to four significant figures.
   (2 marks)

   \includegraphics[max width=\textwidth, alt={}]{b8614dd6-2197-40c3-a673-5bef3e3653a5-2_142_116_2560_157}

   \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\)

1 The curve \(y = x ^ { 3 } - x - 7\) intersects the \(x\)-axis at the point where \(x = \alpha\).

1. Show that \(\alpha\) lies between 2.0 and 2.1.
2. Show that the equation \(x ^ { 3 } - x - 7 = 0\) can be rearranged in the form \(x = \sqrt [ 3 ] { x + 7 }\).
3. Use the iteration \(x _ { n + 1 } = \sqrt [ 3 ] { x _ { n } + 7 }\) with \(x _ { 1 } = 2\) to find the values of \(x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving your answers to three significant figures.

1

1. The curve with equation $$y = \frac { \cos x } { 2 x + 1 } , \quad x > - \frac { 1 } { 2 }$$ intersects the line \(y = \frac { 1 } { 2 }\) at the point where \(x = \alpha\).
   1. Show that \(\alpha\) lies between 0 and \(\frac { \pi } { 2 }\).
   2. Show that the equation \(\frac { \cos x } { 2 x + 1 } = \frac { 1 } { 2 }\) can be rearranged into the form $$x = \cos x - \frac { 1 } { 2 }$$
   3. Use the iteration \(x _ { n + 1 } = \cos x _ { n } - \frac { 1 } { 2 }\) with \(x _ { 1 } = 0\) to find \(x _ { 3 }\), giving your answer to three decimal places.
2. 1. Given that \(y = \frac { \cos x } { 2 x + 1 }\), use the quotient rule to find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
   2. Hence find the gradient of the normal to the curve \(y = \frac { \cos x } { 2 x + 1 }\) at the point on the curve where \(x = 0\).

1 The curve \(y = 3 ^ { x }\) intersects the curve \(y = 10 - x ^ { 3 }\) at the point where \(x = \alpha\).

1. Show that \(\alpha\) lies between 1 and 2 .
2. 1. Show that the equation \(3 ^ { x } = 10 - x ^ { 3 }\) can be rearranged into the form $$x = \sqrt [ 3 ] { 10 - 3 ^ { x } }$$
   2. Use the iteration \(x _ { n + 1 } = \sqrt [ 3 ] { 10 - 3 ^ { x _ { n } } }\) with \(x _ { 1 } = 1\) to find the values of \(x _ { 2 }\) and \(x _ { 3 }\), giving your answers to three decimal places.

2 For \(0 < x \leqslant 2\), the curves with equations \(y = 4 \ln x\) and \(y = \sqrt { x }\) intersect at a single point where \(x = \alpha\).

1. Show that \(\alpha\) lies between 0.5 and 1.5.
2. Show that the equation \(4 \ln x = \sqrt { x }\) can be rearranged into the form $$x = \mathrm { e } ^ { \left( \frac { \sqrt { x } } { 4 } \right) }$$
3. Use the iterative formula $$x _ { n + 1 } = \mathrm { e } ^ { \left( \frac { \sqrt { x _ { n } } } { 4 } \right) }$$ with \(x _ { 1 } = 0.5\) to find the values of \(x _ { 2 }\) and \(x _ { 3 }\), giving your answers to three decimal places.
4. Figure 1, on the page 3, shows a sketch of parts of the graphs of \(y = \mathrm { e } ^ { \left( \frac { \sqrt { x } } { 4 } \right) }\) 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. \begin{figure}[h]
   \captionsetup{labelformat=empty} \caption{Figure 1} \includegraphics[alt={},max width=\textwidth]{d3c66c34-b09c-4223-8383-cf0a68419bf9-3_1285_1543_356_296}
   \end{figure}

3

1. The equation \(\mathrm { e } ^ { - x } - 2 + \sqrt { x } = 0\) has a single root, \(\alpha\).
   Show that \(\alpha\) lies between 3 and 4 .
2. Use the recurrence relation \(x _ { n + 1 } = \left( 2 - e ^ { - x _ { n } } \right) ^ { 2 }\), with \(x _ { 1 } = 3.5\), to find \(x _ { 2 }\) and \(x _ { 3 }\), giving your answers to three decimal places.
3. The diagram below shows parts of the graphs of \(y = \left( 2 - \mathrm { e } ^ { - x } \right) ^ { 2 }\) and \(y = x\), and a position of \(x _ { 1 }\). On the diagram, draw a staircase or cobweb diagram to show how convergence takes place, indicating the positions of \(x _ { 2 }\) and \(x _ { 3 }\) on the \(x\)-axis. \includegraphics[max width=\textwidth, alt={}, center]{063bbfa5-df49-44a1-8143-5e076397f63f-03_1100_1402_881_367}

2 A curve has equation \(y = 2 \ln ( 2 \mathrm { e } - x )\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
2. Find an equation of the normal to the curve \(y = 2 \ln ( 2 \mathrm { e } - x )\) at the point on the curve where \(x = \mathrm { e }\).
   [0pt] [4 marks]
3. The curve \(y = 2 \ln ( 2 \mathrm { e } - x )\) intersects the line \(y = x\) at a single point, where \(x = \alpha\).
   1. Show that \(\alpha\) lies between 1 and 3 .
   2. Use the recurrence relation $$x _ { n + 1 } = 2 \ln \left( 2 \mathrm { e } - x _ { n } \right)$$ with \(x _ { 1 } = 1\) to find the values of \(x _ { 2 }\) and \(x _ { 3 }\), giving your answers to three decimal places.
   3. Figure 1, on the opposite page, shows a sketch of parts of the graphs of \(y = 2 \ln ( 2 \mathrm { e } - x )\) 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.
      [0pt] [2 marks] \section*{(c)(iii)} \begin{figure}[h]
      \captionsetup{labelformat=empty} \caption{Figure 1} \includegraphics[alt={},max width=\textwidth]{57412ec0-ad97-4418-8ba8-93f1f7d8aac1-05_864_1284_1802_386}
      \end{figure}

2 The curve with equation \(y = x ^ { x }\), where \(x > 0\), intersects the line \(y = 5\) at a single point, where \(x = \alpha\).

1. Show that \(\alpha\) lies between 2 and 3 .
2. Show that the equation \(x ^ { x } = 5\) can be rearranged into the form $$x = \mathrm { e } ^ { \left( \frac { \ln 5 } { x } \right) }$$
3. Use the iterative formula $$x _ { n + 1 } = \mathrm { e } ^ { \left( \frac { \ln 5 } { x _ { n } } \right) }$$ with \(x _ { 1 } = 2\) to find the values of \(x _ { 2 }\) and \(x _ { 3 }\), giving your answers to three decimal places.
4. 1. Use Simpson's rule with 7 ordinates ( 6 strips) to find an approximation to $$\int _ { 0.5 } ^ { 1.7 } \left( 5 - x ^ { x } \right) \mathrm { d } x$$ giving your answer to three significant figures.
   2. Hence find an approximation to \(\int _ { 0.5 } ^ { 1.7 } x ^ { x } \mathrm {~d} x\).

3. $$f ( x ) = x ^ { 2 } + 5 x - 2 \sec x , \quad x \in \mathbb { R } , \quad - \frac { \pi } { 2 } < x < \frac { \pi } { 2 } .$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root in the interval [1,1.5]. A more accurate estimate of this root is to be found using iterations of the form $$x _ { n + 1 } = \arccos \mathrm { g } \left( x _ { n } \right) .$$
2. Find a suitable form for \(\mathrm { g } ( x )\) and use this formula with \(x _ { 0 } = 1.25\) to find \(x _ { 1 } , x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\). Give the value of \(x _ { 4 }\) to 3 decimal places. The curve \(y = \mathrm { f } ( x )\) has a stationary point at \(P\).
3. Show that the \(x\)-coordinate of \(P\) is 1.0535 correct to 5 significant figures.

8. A curve has the equation \(y = x ^ { 2 } - \sqrt { 4 + \ln x }\).

1. Show that the tangent to the curve at the point where \(x = 1\) has the equation $$7 x - 4 y = 11$$ The curve has a stationary point with \(x\)-coordinate \(\alpha\).
2. Show that \(0.3 < \alpha < 0.4\)
3. Show that \(\alpha\) is a solution of the equation $$x = \frac { 1 } { 2 } ( 4 + \ln x ) ^ { - \frac { 1 } { 4 } }$$
4. Use the iteration formula $$x _ { n + 1 } = \frac { 1 } { 2 } \left( 4 + \ln x _ { n } \right) ^ { - \frac { 1 } { 4 } }$$ with \(x _ { 0 } = 0.35\), to find \(x _ { 1 } , x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving your answers to 5 decimal places. END

7 \includegraphics[max width=\textwidth, alt={}, center]{43ed8ec7-67f1-418a-8d4e-ee96448647fd-4_351_314_255_861} One end of a light elastic string, of natural length \(\frac { 2 } { 3 } R \mathrm {~m}\) and with modulus of elasticity 1.2 mgN , is attached to the highest point \(A\) of a smooth fixed sphere with centre \(O\) and radius \(R \mathrm {~m}\). A particle \(P\) of mass \(m \mathrm {~kg}\) is attached to the other end of the string and is in contact with the surface of the sphere, where the angle \(A O P\) is equal to \(\theta\) radians (see diagram).

1. Given that \(P\) is in equilibrium at the point where \(\theta = \alpha\), show that \(1.8 \alpha - \sin \alpha - 1.2 = 0\). Hence show that \(\alpha = 1.18\) correct to 3 significant figures. \(P\) is now released from rest at the point of the surface of the sphere where \(\theta = \frac { 2 } { 3 }\), and starts to move downwards on the surface. For an instant when \(\theta = \alpha\),
2. state the direction of the acceleration of \(P\),
3. find the magnitude of the acceleration of \(P\).

1

1. Show that the equation $$x ^ { 3 } + 2 x - 2 = 0$$ has a root between 0.5 and 1 .
2. Use linear interpolation once to find an estimate of this root. Give your answer to two decimal places.

5 The diagram below (not to scale) shows a part of a curve \(y = \mathrm { f } ( x )\) which passes through the points \(A ( 2 , - 10 )\) and \(B ( 5,22 )\).

1. 1. On the diagram, draw a line which illustrates the method of linear interpolation for solving the equation \(\mathrm { f } ( x ) = 0\). The point of intersection of this line with the \(x\)-axis should be labelled \(P\).
   2. Calculate the \(x\)-coordinate of \(P\). Give your answer to one decimal place.
2. 1. On the same diagram, draw a line which illustrates the Newton-Raphson method for solving the equation \(\mathrm { f } ( x ) = 0\), with initial value \(x _ { 1 } = 2\). The point of intersection of this line with the \(x\)-axis should be labelled \(Q\).
   2. Given that the gradient of the curve at \(A\) is 8 , calculate the \(x\)-coordinate of \(Q\). Give your answer as an exact decimal. \includegraphics[max width=\textwidth, alt={}, center]{f9345653-d426-4350-bf1d-901506211078-3_876_1063_1779_523}

2

1. Show that the equation $$x ^ { 3 } + x - 7 = 0$$ has a root between 1.6 and 1.8.
2. Use interval bisection twice, starting with the interval in part (a), to give this root to one decimal place.

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)

7

1. Show that the equation \(4 x ^ { 3 } - x - 540000 = 0\) has a root, \(\alpha\), in the interval \(51 < \alpha < 52\).
2. It is given that \(S _ { n } = \sum _ { r = 1 } ^ { n } ( 2 r - 1 ) ^ { 2 }\).
   1. Use the formulae for \(\sum _ { r = 1 } ^ { n } r ^ { 2 }\) and \(\sum _ { r = 1 } ^ { n } r\) to show that \(S _ { n } = \frac { n } { 3 } \left( k n ^ { 2 } - 1 \right)\), where \(k\) is an integer to be found.
   2. Hence show that \(6 S _ { n }\) can be written as the product of three consecutive integers.
3. Find the smallest value of \(N\) for which the sum of the squares of the first \(N\) odd numbers is greater than 180000 .

7

1. The equation \(2 x ^ { 3 } + 5 x ^ { 2 } + 3 x - 132000 = 0\) has exactly one real root \(\alpha\).
   1. Show that \(\alpha\) lies in the interval \(39 < \alpha < 40\).
   2. Taking \(x _ { 1 } = 40\) as a first approximation to \(\alpha\), use the Newton-Raphson method to find a second approximation, \(x _ { 2 }\), to \(\alpha\). Give your answer to two decimal places.
2. Use the formulae for \(\sum _ { r = 1 } ^ { n } r ^ { 2 }\) and \(\sum _ { r = 1 } ^ { n } r\) to show that $$\sum _ { r = 1 } ^ { n } 2 r ( 3 r + 2 ) = n ( n + p ) ( 2 n + q )$$ where \(p\) and \(q\) are integers.
3. 1. Express \(\log _ { 8 } 4 ^ { r }\) in the form \(\lambda r\), where \(\lambda\) is a rational number.
   2. By first finding a suitable cubic inequality for \(k\), find the greatest value of \(k\) for which \(\sum _ { r = k + 1 } ^ { 60 } ( 3 r + 2 ) \log _ { 8 } 4 ^ { r }\) is greater than 106060.
      [0pt] [4 marks]

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.

5 The root of the equation \(\mathrm { f } ( x ) = 0\) is being found using the method of interval bisection. Some of the associated spreadsheet output is shown in the table below. The formula in cell B2 is \(\quad = \mathrm { EXP } ( \mathrm { A } 2 ) - \mathrm { A } 2 ^ { \wedge } 2 - \mathrm { A } 2 - 2\).

1. Write down the equation whose root is being found.
2. Write down a suitable formula for cell E2. The formula in cell A3 is $$= \mathrm { IF } ( \mathrm {~F} 2 < 0 , \mathrm { E } 2 , \mathrm {~A} 2 )$$ .
3. Write down a similar formula for cell C3.
4. Complete row 6 of the table on the copy in the Printed Answer Booklet.
5. Without doing any calculations, write down the value of the root correct to the number of decimal places which seems justified. You must explain the precision quoted.
6. Determine how many more applications of the bisection method are needed such that the interval which contains the root is less than 0.0005 .

6 Fig. 6.1 shows the graph of \(y = \mathrm { e } ^ { 3 x } - 11 x - 0.5\) for \(- 0.5 \leqslant x \leqslant 1\). \begin{figure}[h] \end{figure} The equation \(\mathrm { e } ^ { 3 x } - 11 x - 0.5 = 0\) has two roots, \(\alpha\) and \(\beta\), such that \(\alpha < \beta\). Dennis is going to use the method of interval bisection with starting values denoted by \(a\) and \(b\).

1. Explain why the method of interval bisection starting with \(a = 0\) and \(b = 1\) may not be used to find either \(\alpha\) or \(\beta\). Dennis uses the method of interval bisection starting with \(a = 0\) and \(b = 0.5\) to find \(\alpha\). Some spreadsheet output is shown in Fig. 6.2. \begin{table}[h]

   	A	B	C	D	E	F
   1	a	f(a)	b	f(b)	\(x _ { \text {new } }\)	\(\mathrm { f } \left( x _ { \text {new } } \right)\)
   2	0	0.5	0.5	-1.51831	0.25	-1.133
   3	0	0.5	0.25	-1.133	0.125	-0.42
   4	0	0.5	0.125	-0.42001	0.0625	0.01873
   5	0.0625	0.01873	0.125	-0.42001	0.09375	-0.2065

   \captionsetup{labelformat=empty} \caption{Fig. 6.2}
   \end{table} Dennis states that the formula in cell B2 is $$= \operatorname { EXP } \left( 3 ^ { * } \mathrm {~A} 1 \right) - 11 \mathrm {~A} 2 - 0.5$$ Dennis has made two errors.
2. Write a correct version of Dennis's formula for cell B2. The formula in cell A3, which is correct, is
   = IF(F2 > 0, E2, A2)
3. Write a suitable formula for cell C3.
4. Use the information in Fig. 6.2 to
   Liren uses a different method to find a sequence of estimates of the value of \(\beta\) using a spreadsheet. The output, together with some further analysis, is shown in Fig. 6.3. \begin{table}[h]

   	A	B	C	D
   1	\(x\)	f(x)	difference	ratio
   2	0.4	-1.5799
   3	0.6	-1.0504
   4	0.99671	8.4245
   5	0.64398	-0.6809	-0.35273
   6	0.67036	-0.4026	0.026378	-0.0748
   7	0.70852	0.08386	0.038164	1.44682
   8	0.70194	-0.0075	-0.00658	-0.1724
   9	0.70248	-0.0001	0.00054	-0.082
   10	0.70249	\(1.8 \mathrm { E } - 07\)	\(8.88 \mathrm { E } - 06\)	0.01646

   \captionsetup{labelformat=empty} \caption{Fig. 6.3}
   \end{table} The formula in cell A4 is $$= ( \mathrm { A } 2 * \mathrm {~B} 3 - \mathrm { A } 3 * \mathrm {~B} 2 ) / ( \mathrm { B } 3 - \mathrm { B } 2 )$$
5. State the method being used.
6. Explain what the values in column D tell you about the order of convergence of this sequence of estimates. Liren states that \(\beta = 0.70249\) correct to 5 decimal places.
7. Determine whether Liren is correct.

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?