Sign Change with Function Evaluation

5. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the curve with equation $$y = 6 \ln ( 2 x + 3 ) - \frac { 1 } { 2 } x ^ { 2 } + 4 \quad x > - \frac { 3 } { 2 }$$ The curve cuts the negative \(x\)-axis at the point \(P\), as shown in Figure 1.

1. Show that the \(x\) coordinate of \(P\) lies in the interval \([ - 1.25 , - 1.2 ]\) The curve cuts the positive \(x\)-axis at the point \(Q\), also shown in Figure 1.
   Using the iterative formula $$x _ { n + 1 } = \sqrt { 12 \ln \left( 2 x _ { n } + 3 \right) + 8 } \text { with } x _ { 1 } = 6$$
2. 1. find, to 4 decimal places, the value of \(x _ { 2 }\)
   2. find, by continued iteration, the \(x\) coordinate of \(Q\). Give your answer to 4 decimal places. The curve has a maximum turning point at \(M\), as shown in Figure 1.
3. Using calculus and showing each stage of your working, find the \(x\) coordinate of \(M\).

7. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of part of the curve with equation \(y = \mathrm { f } ( x )\) where $$\mathrm { f } ( x ) = \left( x ^ { 2 } + 3 x + 1 \right) \mathrm { e } ^ { x ^ { 2 } }$$ The curve cuts the \(x\)-axis at points \(A\) and \(B\) as shown in Figure 2 .

1. Calculate the \(x\) coordinate of \(A\) and the \(x\) coordinate of \(B\), giving your answers to 3 decimal places.
2. Find \(\mathrm { f } ^ { \prime } ( x )\). The curve has a minimum turning point at the point \(P\) as shown in Figure 2.
3. Show that the \(x\) coordinate of \(P\) is the solution of $$x = - \frac { 3 \left( 2 x ^ { 2 } + 1 \right) } { 2 \left( x ^ { 2 } + 2 \right) }$$
4. Use the iteration formula $$x _ { n + 1 } = - \frac { 3 \left( 2 x _ { n } ^ { 2 } + 1 \right) } { 2 \left( x _ { n } ^ { 2 } + 2 \right) } , \quad \text { with } x _ { 0 } = - 2.4$$ to calculate the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\), giving your answers to 3 decimal places. The \(x\) coordinate of \(P\) is \(\alpha\).
5. By choosing a suitable interval, prove that \(\alpha = - 2.43\) to 2 decimal places.

6. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of part of the curve with equation $$y = 2 \cos \left( \frac { 1 } { 2 } x ^ { 2 } \right) + x ^ { 3 } - 3 x - 2$$ The curve crosses the \(x\)-axis at the point \(Q\) and has a minimum turning point at \(R\).

1. Show that the \(x\) coordinate of \(Q\) lies between 2.1 and 2.2
2. Show that the \(x\) coordinate of \(R\) is a solution of the equation $$x = \sqrt { 1 + \frac { 2 } { 3 } x \sin \left( \frac { 1 } { 2 } x ^ { 2 } \right) }$$ Using the iterative formula $$x _ { n + 1 } = \sqrt { 1 + \frac { 2 } { 3 } x _ { n } \sin \left( \frac { 1 } { 2 } x _ { n } ^ { 2 } \right) } , \quad x _ { 0 } = 1.3$$
3. find the values of \(x _ { 1 }\) and \(x _ { 2 }\) to 3 decimal places.

8 \includegraphics[max width=\textwidth, alt={}, center]{e0e2a26b-d4d6-46ea-ac12-a882f3465e5e-3_588_915_954_614} The diagram shows part of each of the curves \(y = e ^ { \frac { 1 } { 5 } x }\) and \(y = \sqrt [ 3 ] { } ( 3 x + 8 )\). The curves meet, as shown in the diagram, at the point \(P\). The region \(R\), shaded in the diagram, is bounded by the two curves and by the \(y\)-axis.

1. Show by calculation that the \(x\)-coordinate of \(P\) lies between 5.2 and 5.3.
2. Show that the \(x\)-coordinate of \(P\) satisfies the equation \(x = \frac { 5 } { 3 } \ln ( 3 x + 8 )\).
3. Use an iterative formula, based on the equation in part (ii), to find the \(x\)-coordinate of \(P\) correct to 2 decimal places.
4. Use integration, and your answer to part (iii), to find an approximate value of the area of the region \(R\). \includegraphics[max width=\textwidth, alt={}, center]{e0e2a26b-d4d6-46ea-ac12-a882f3465e5e-4_625_647_264_749} The function f is defined by \(\mathrm { f } ( x ) = \sqrt { } ( m x + 7 ) - 4\), where \(x \geqslant - \frac { 7 } { m }\) and \(m\) is a positive constant. The diagram shows the curve \(y = \mathrm { f } ( x )\).
5. A sequence of transformations maps the curve \(y = \sqrt { } x\) to the curve \(y = \mathrm { f } ( x )\). Give details of these transformations.
6. Explain how you can tell that f is a one-one function and find an expression for \(\mathrm { f } ^ { - 1 } ( x )\).
7. It is given that the curves \(y = \mathrm { f } ( x )\) and \(y = \mathrm { f } ^ { - 1 } ( x )\) do not meet. Explain how it can be deduced that neither curve meets the line \(y = x\), and hence determine the set of possible values of \(m\). [5]

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 .

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 .