1.09b Sign change methods: understand failure cases

9. \begin{figure}[h] \end{figure} Figure 3 shows a sketch of part of the curve \(C\) with equation $$y = \sqrt { 3 + 4 \mathrm { e } ^ { x ^ { 2 } } } \quad x \geqslant 0$$

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\), giving your answer in simplest form. The point \(P\) with \(x\) coordinate \(\alpha\) lies on \(C\).
   Given that the tangent to \(C\) at \(P\) passes through the origin, as shown in Figure 3,
2. show that \(x = \alpha\) is a solution of the equation $$4 x ^ { 2 } e ^ { x ^ { 2 } } - 4 e ^ { x ^ { 2 } } - 3 = 0$$
3. Hence show that \(\alpha\) lies between 1 and 2
4. Show that the equation in part (b) can be written in the form $$x = \frac { 1 } { 2 } \sqrt { 4 + 3 \mathrm { e } ^ { - x ^ { 2 } } }$$ The iteration formula $$x _ { n + 1 } = \frac { 1 } { 2 } \sqrt { 4 + 3 \mathrm { e } ^ { - x _ { n } ^ { 2 } } }$$ with \(x _ { 1 } = 1\) is used to find an approximation for \(\alpha\).
5. Use the iteration formula to find, to 4 decimal places, the value of
   1. \(X _ { 3 }\)
   2. \(\alpha\)

5. $$f ( x ) = - x ^ { 3 } + 4 x ^ { 2 } - 6$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root between \(x = 1\) and \(x = 2\)
2. Show that the equation \(\mathrm { f } ( x ) = 0\) can be rewritten as $$x = \sqrt { \left( \frac { 6 } { 4 - x } \right) }$$
3. Starting with \(x _ { 1 } = 1.5\) use the iteration \(x _ { n + 1 } = \sqrt { \left( \frac { 6 } { 4 - x _ { n } } \right) }\) to calculate the values of \(x _ { 2 }\), \(x _ { 3 }\) and \(x _ { 4 }\) giving all your answers to 4 decimal places.
4. Using a suitable interval, show that 1.572 is a root of \(\mathrm { f } ( x ) = 0\) correct to 3 decimal places.

2. $$f ( x ) = x ^ { 3 } - 5 x + 16$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) can be rewritten as $$x = ( a x + b ) ^ { \frac { 1 } { 3 } }$$ giving the values of the constants \(a\) and \(b\). The equation \(\mathrm { f } ( x ) = 0\) has exactly one real root \(\alpha\), where \(\alpha = - 3\) to one significant figure.
2. Starting with \(x _ { 1 } = - 3\), use the iteration $$x _ { n + 1 } = \left( a x _ { n } + b \right) ^ { \frac { 1 } { 3 } }$$ with the values of \(a\) and \(b\) found in part (a), to calculate the values of \(x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving all your answers to 3 decimal places.
3. Using a suitable interval, show that \(\alpha = - 3.17\) correct to 2 decimal places.

3. $$f ( x ) = 2 ^ { x - 1 } - 4 + 1.5 x \quad x \in \mathbb { R }$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) can be written as $$x = \frac { 1 } { 3 } \left( 8 - 2 ^ { x } \right)$$ The equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\), where \(\alpha = 1.6\) to one decimal place.
2. Starting with \(x _ { 0 } = 1.6\), use the iteration formula $$x _ { n + 1 } = \frac { 1 } { 3 } \left( 8 - 2 ^ { x _ { n } } \right)$$ to calculate the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\), giving your answers to 3 decimal places.
3. By choosing a suitable interval, prove that \(\alpha = 1.633\) to 3 decimal places.

1. $$f ( x ) = 2 x ^ { 3 } + x - 20$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) can be rewritten as $$x = \sqrt [ 3 ] { a - b x }$$ where \(a\) and \(b\) are positive constants to be determined.
2. Starting with \(x _ { 1 } = 2.1\) use the iteration formula \(x _ { n + 1 } = \sqrt [ 3 ] { a - b x _ { n } }\), with the numerical values of \(a\) and \(b\), to calculate the values of \(x _ { 2 }\) and \(x _ { 3 }\) giving your answers to 3 decimal places.
3. Using a suitable interval, show that 2.077 is a root of the equation \(\mathrm { f } ( x ) = 0\) correct to 3 decimal places.
4. Hence state a root, to 3 decimal places, of the equation $$2 ( x + 2 ) ^ { 3 } + x - 18 = 0$$

   VIIIV SIHI NI JIIYM ION OC

   VIIV SIHI NI JIIIM ION OC

   VIIV SIHI NI JIIYM ION OC

3. $$\mathrm { f } ( x ) = \ln ( x + 2 ) - x + 1 , \quad x > - 2 , x \in \mathbb { R } .$$

1. Show that there is a root of \(\mathrm { f } ( x ) = 0\) in the interval \(2 < x < 3\).
2. Use the iterative formula $$x _ { n + 1 } = \ln \left( x _ { n } + 2 \right) + 1 , x _ { 0 } = 2.5$$ to calculate the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\) giving your answers to 5 decimal places.
3. Show that \(x = 2.505\) is a root of \(\mathrm { f } ( x ) = 0\) correct to 3 decimal places.

7. $$f ( x ) = 3 x e ^ { x } - 1$$ The curve with equation \(y = \mathrm { f } ( x )\) has a turning point \(P\).

1. Find the exact coordinates of \(P\). The equation \(\mathrm { f } ( x ) = 0\) has a root between \(x = 0.25\) and \(x = 0.3\)
2. Use the iterative formula $$x _ { n + 1 } = \frac { 1 } { 3 } \mathrm { e } ^ { - x _ { n } }$$ with \(x _ { 0 } = 0.25\) to find, to 4 decimal places, the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\).
3. By choosing a suitable interval, show that a root of \(\mathrm { f } ( x ) = 0\) is \(x = 0.2576\) correct to 4 decimal places.

2. $$f ( x ) = x ^ { 3 } + 2 x ^ { 2 } - 3 x - 11$$

1. Show that \(\mathrm { f } ( x ) = 0\) can be rearranged as $$x = \sqrt { } \left( \frac { 3 x + 11 } { x + 2 } \right) , \quad x \neq - 2 .$$ The equation \(\mathrm { f } ( x ) = 0\) has one positive root \(\alpha\). The iterative formula \(x _ { n + 1 } = \sqrt { } \left( \frac { 3 x _ { n } + 11 } { x _ { n } + 2 } \right)\) is used to find an approximation to \(\alpha\).
2. Taking \(x _ { 1 } = 0\), find, to 3 decimal places, the values of \(x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\).
3. Show that \(\alpha = 2.057\) correct to 3 decimal places.

3. \(\mathrm { f } ( x ) = 4 \operatorname { cosec } x - 4 x + 1\), where \(x\) is in radians.

1. Show that there is a root \(\alpha\) of \(\mathrm { f } ( x ) = 0\) in the interval [1.2,1.3].
2. Show that the equation \(\mathrm { f } ( x ) = 0\) can be written in the form $$x = \frac { 1 } { \sin x } + \frac { 1 } { 4 }$$
3. Use the iterative formula $$x _ { n + 1 } = \frac { 1 } { \sin x _ { n } } + \frac { 1 } { 4 } , \quad x _ { 0 } = 1.25$$ to calculate the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\), giving your answers to 4 decimal places.
4. By considering the change of sign of \(\mathrm { f } ( x )\) in a suitable interval, verify that \(\alpha = 1.291\) correct to 3 decimal places.

4. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the curve with equation \(y = g ( x )\), where $$\mathrm { g } ( x ) = \left| 4 \mathrm { e } ^ { 2 x } - 25 \right| , \quad x \in \mathbb { R }$$ The curve cuts the \(y\)-axis at the point \(A\) and meets the \(x\)-axis at the point \(B\). The curve has an asymptote \(y = k\), where \(k\) is a constant, as shown in Figure 1

1. Find, giving each answer in its simplest form,
   1. the \(y\) coordinate of the point \(A\),
   2. the exact \(x\) coordinate of the point \(B\),
   3. the value of the constant \(k\). The equation \(\mathrm { g } ( x ) = 2 x + 43\) has a positive root at \(x = \alpha\)
2. Show that \(\alpha\) is a solution of \(x = \frac { 1 } { 2 } \ln \left( \frac { 1 } { 2 } x + 17 \right)\) The iteration formula $$x _ { n + 1 } = \frac { 1 } { 2 } \ln \left( \frac { 1 } { 2 } x _ { n } + 17 \right)$$ can be used to find an approximation for \(\alpha\)
3. Taking \(x _ { 0 } = 1.4\) find the values of \(x _ { 1 }\) and \(x _ { 2 }\) Give each answer to 4 decimal places.
4. By choosing a suitable interval, show that \(\alpha = 1.437\) to 3 decimal places. \includegraphics[max width=\textwidth, alt={}, center]{d3ba2776-eedb-48f0-834f-41aa454afba3-07_2258_47_315_37}

$$f ( x ) = x - 4 - \cos ( 5 \sqrt { x } ) \quad x > 0$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) in the interval [2.5, 3.5]
   [0pt]
2. Use linear interpolation once on the interval [2.5, 3.5] to find an approximation to \(\alpha\), giving your answer to 2 decimal places.
   (ii) $$\operatorname { g } ( x ) = \frac { 1 } { 10 } x ^ { 2 } - \frac { 1 } { 2 x ^ { 2 } } + x - 11 \quad x > 0$$
3. Determine \(\mathrm { g } ^ { \prime } ( x )\). The equation \(\mathrm { g } ( x ) = 0\) has a root \(\beta\) in the interval [6,7]
4. Using \(x _ { 0 } = 6\) as a first approximation to \(\beta\), apply the Newton-Raphson procedure once to \(\mathrm { g } ( x )\) to find a second approximation to \(\beta\), giving your answer to 3 decimal places.

4. $$\mathrm { f } ( x ) = x ^ { \frac { 3 } { 2 } } - 3 x ^ { \frac { 1 } { 2 } } - 3 , \quad x > 0$$ Given that \(\alpha\) is the only real root of the equation \(\mathrm { f } ( x ) = 0\),

1. show that \(4 < \alpha < 5\)
2. Taking 4.5 as a first approximation to \(\alpha\), apply the Newton-Raphson procedure once to \(\mathrm { f } ( x )\) to find a second approximation to \(\alpha\), giving your answer to 3 decimal places.
   [0pt]
3. Use linear interpolation once on the interval [4,5] to find another approximation to \(\alpha\), giving your answer to 3 decimal places.

7. $$f ( x ) = x ^ { \frac { 3 } { 2 } } + x - 3$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root, \(\alpha\), in the interval \([ 1,2 ]\) [0pt]
2. Starting with the interval [1, 2], use interval bisection twice to show that \(\alpha\) lies in the interval [1.25, 1.5]
3. 1. Determine \(\mathrm { f } ^ { \prime } ( x )\)
   2. Using 1.375 as a first approximation for \(\alpha\), apply the Newton-Raphson process once to \(\mathrm { f } ( x )\) to determine a second approximation for \(\alpha\), giving your answer to 3 decimal places.
      [0pt]
4. Use linear interpolation once on the interval [1.25,1.5] to obtain a different approximation for \(\alpha\), giving your answer to 3 decimal places.

2. $$f ( x ) = 3 x ^ { 2 } - \frac { 11 } { x ^ { 2 } }$$

1. Write down, to 3 decimal places, the value of \(\mathrm { f } ( 1.3 )\) and the value of \(\mathrm { f } ( 1.4 )\). The equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) between 1.3 and 1.4
   [0pt]
2. Starting with the interval [1.3, 1.4], use interval bisection to find an interval of width 0.025 which contains \(\alpha\).
3. Taking 1.4 as a first approximation to \(\alpha\), apply the Newton-Raphson procedure once to \(\mathrm { f } ( x )\) to obtain a second approximation to \(\alpha\), giving your answer to 3 decimal places.

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.

3. $$\mathrm { f } ( x ) = x ^ { 3 } - \frac { 7 } { x } + 2 , \quad x > 0$$

1. Show that \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) between 1.4 and 1.5
   [0pt]
2. Starting with the interval [1.4,1.5], use interval bisection twice to find an interval of width 0.025 that contains \(\alpha\).
3. Taking 1.45 as a first approximation to \(\alpha\), apply the Newton-Raphson procedure once to \(\mathrm { f } ( x ) = x ^ { 3 } - \frac { 7 } { x } + 2\) to obtain a second approximation to \(\alpha\), giving your answer to 3 decimal places.

1. $$f ( x ) = 3 ^ { x } + 3 x - 7$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) between \(x = 1\) and \(x = 2\).
2. Starting with the interval \([ 1,2 ]\), use interval bisection twice to find an interval of width 0.25 which contains \(\alpha\).

8. $$f ( x ) = x ^ { 3 } - 2 x - 3$$

1. Show that \(\mathrm { f } ( x ) = 0\) has a root, \(\alpha\), in the interval \([ 1,2 ]\).
2. Starting with the interval \([ 1,2 ]\), use interval bisection twice to find an interval of width 0.25 which contains \(\alpha\).
3. Using \(x _ { 0 } = 1.8\) as a first approximation to \(\alpha\), apply the Newton-Raphson procedure once to \(\mathrm { f } ( x )\) to find a second approximation to \(\alpha\), giving your answer to 3 significant figures.

2. $$\mathrm { f } ( x ) = \cos \left( x ^ { 2 } \right) - x + 3 , \quad 0 < x < \pi$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) in the interval \([ 2.5,3 ]\).
   [0pt]
2. Use linear interpolation once on the interval [2.5,3] to find an approximation for \(\alpha\), giving your answer to 2 decimal places.

2. $$\mathrm { f } ( x ) = 3 \cos 2 x + x - 2 , \quad - \pi \leqslant x < \pi$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) in the interval [2,3].
   [0pt]
2. Use linear interpolation once on the interval [2,3] to find an approximation to \(\alpha\). Give your answer to 3 decimal places.
3. The equation \(\mathrm { f } ( x ) = 0\) has another root \(\beta\) in the interval \([ - 1,0 ]\). Starting with this interval, use interval bisection to find an interval of width 0.25 which contains \(\beta\).

2. $$\mathrm { f } ( x ) = x ^ { 3 } - \frac { 5 } { 2 x ^ { \frac { 3 } { 2 } } } + 2 x - 3 , \quad x > 0$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) in the interval [1.1,1.5].
2. Find f'(x).
3. Using \(x _ { 0 } = 1.1\) as a first approximation to \(\alpha\), apply the Newton-Raphson procedure once to \(\mathrm { f } ( x )\) to find a second approximation to \(\alpha\), giving your answer to 3 decimal places.

2. In the interval \(13 < x < 14\), the equation $$3 + x \sin \left( \frac { x } { 4 } \right) = 0 , \text { where } x \text { is measured in radians, }$$ has exactly one root, \(\alpha\).
[0pt]

1. Starting with the interval [13,14], use interval bisection twice to find an interval of width 0.25 which contains \(\alpha\).
   [0pt]
2. Use linear interpolation once on the interval [13,14] to find an approximate value for \(\alpha\). Give your answer to 3 decimal places.

1. 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 iterative 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.

9 The managers of a car breakdown recovery service are discussing whether the number of breakdowns per day can be modelled by a Poisson distribution. They agree that breakdowns occur randomly. Manager \(A\) says, "it must be assumed that breakdowns occur at a constant rate throughout the day".

1. Give an improved version of Manager \(A\) 's statement, and explain why the improvement is necessary.
2. Explain whether you think your improved statement is likely to hold in this context. Assume now that the number \(B\) of breakdowns per day can be modelled by the distribution \(\operatorname { Po } ( \lambda )\).
3. Given that \(\lambda = 9.0\) and \(\mathrm { P } \left( B > B _ { 0 } \right) < 0.1\), use tables to find the smallest possible value of \(B _ { 0 }\), and state the corresponding value of \(\mathrm { P } \left( B > B _ { 0 } \right)\).
4. Given that \(\mathrm { P } ( B = 2 ) = 0.0072\), show that \(\lambda\) satisfies an equation of the form \(\lambda = 0.12 \mathrm { e } ^ { k \lambda }\), for a value of \(k\) to be stated. Evaluate the expression \(0.12 \mathrm { e } ^ { k \lambda }\) for \(\lambda = 8.5\) and \(\lambda = 8.6\), giving your answers correct to 4 decimal places. What can be deduced about a possible value of \(\lambda\) ?

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.