Derive stationary point equation

4. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the curve with equation \(y = 8 x - x \mathrm { e } ^ { 3 x } , x \geqslant 0\) The curve meets the \(x\)-axis at the origin and cuts the \(x\)-axis at the point \(A\).

1. Find the exact \(x\) coordinate of \(A\), giving your answer in its simplest form. The curve has a maximum turning point at the point \(M\).
2. Show, by using calculus, that the \(x\) coordinate of \(M\) is a solution of $$x = \frac { 1 } { 3 } \ln \left( \frac { 8 } { 1 + 3 x } \right)$$
3. Use the iterative formula $$x _ { n + 1 } = \frac { 1 } { 3 } \ln \left( \frac { 8 } { 1 + 3 x _ { n } } \right)$$ with \(x _ { 0 } = 0.4\) to calculate the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\), giving your answers to 3 decimal places.

6. $$f ( x ) = x ^ { 2 } - 3 x + 2 \cos \left( \frac { 1 } { 2 } x \right) , \quad 0 \leqslant x \leqslant \pi$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a solution in the interval \(0.8 < x < 0.9\) The curve with equation \(y = \mathrm { f } ( x )\) has a minimum point \(P\).
2. Show that the \(x\)-coordinate of \(P\) is the solution of the equation $$x = \frac { 3 + \sin \left( \frac { 1 } { 2 } x \right) } { 2 }$$
3. Using the iteration formula $$x _ { n + 1 } = \frac { 3 + \sin \left( \frac { 1 } { 2 } x _ { n } \right) } { 2 } , \quad x _ { 0 } = 2$$ find the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\), giving your answers to 3 decimal places.
4. By choosing a suitable interval, show that the \(x\)-coordinate of \(P\) is 1.9078 correct to 4 decimal places.

4. $$\mathrm { f } ( x ) = 3 \mathrm { e } ^ { x } - \frac { 1 } { 2 } \ln x - 2 , \quad x > 0 .$$

1. Differentiate to find \(\mathrm { f } ^ { \prime } ( x )\). The curve with equation \(y = \mathrm { f } ( x )\) has a turning point at \(P\). The \(x\)-coordinate of \(P\) is \(\alpha\).
2. Show that \(\alpha = \frac { 1 } { 6 } \mathrm { e } ^ { - \alpha }\). The iterative formula $$x _ { n + 1 } = \frac { 1 } { 6 } \mathrm { e } ^ { - x _ { n } } , x _ { 0 } = 1$$ is used to find an approximate value for \(\alpha\).
3. Calculate the values of \(x _ { 1 } , x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving your answers to 4 decimal places.
4. By considering the change of sign of \(\mathrm { f } ^ { \prime } ( x )\) in a suitable interval, prove that \(\alpha = 0.1443\) correct to 4 decimal places.

5. \begin{figure}[h] \end{figure} Figure 2 shows part of the curve with equation $$y = ( 2 x - 1 ) \tan 2 x , \quad 0 \leqslant x < \frac { \pi } { 4 }$$ The curve has a minimum at the point \(P\). The \(x\)-coordinate of \(P\) is \(k\).

1. Show that \(k\) satisfies the equation $$4 k + \sin 4 k - 2 = 0$$ The iterative formula $$x _ { n + 1 } = \frac { 1 } { 4 } \left( 2 - \sin 4 x _ { n } \right) , x _ { 0 } = 0.3$$ is used to find an approximate value for \(k\).
2. Calculate the values of \(x _ { 1 } , x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving your answers to 4 decimal places.
3. Show that \(k = 0.277\), correct to 3 significant figures.

2. A curve \(C\) has equation \(y = \mathrm { e } ^ { 4 x } + x ^ { 4 } + 8 x + 5\)

1. Show that the \(x\) coordinate of any turning point of \(C\) satisfies the equation $$x ^ { 3 } = - 2 - \mathrm { e } ^ { 4 x }$$
2. On the axes given on page 5, sketch, on a single diagram, the curves with equations
   1. \(y = x ^ { 3 }\),
   2. \(y = - 2 - e ^ { 4 x }\) On your diagram give the coordinates of the points where each curve crosses the \(y\)-axis and state the equation of any asymptotes.
3. Explain how your diagram illustrates that the equation \(x ^ { 3 } = - 2 - e ^ { 4 x }\) has only one root. The iteration formula $$x _ { n + 1 } = \left( - 2 - \mathrm { e } ^ { 4 x _ { n } } \right) ^ { \frac { 1 } { 3 } } , \quad x _ { 0 } = - 1$$ can be used to find an approximate value for this root.
4. Calculate the values of \(x _ { 1 }\) and \(x _ { 2 }\), giving your answers to 5 decimal places.
5. Hence deduce the coordinates, to 2 decimal places, of the turning point of the curve \(C\). \includegraphics[max width=\textwidth, alt={}, center]{be00fdaa-2fe3-4f06-a710-08ec67fb911e-04_1285_1294_308_331}

6 The curve with equation \(y = \frac { 3 x + 4 } { x ^ { 3 } - 4 x ^ { 2 } + 2 }\) has a stationary point at \(P\). It is given that \(P\) is close to the point with coordinates \(( 2.4 , - 1.6 )\).

1. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and show that the \(x\)-coordinate of \(P\) satisfies the equation $$x = \sqrt [ 3 ] { \frac { 16 } { 3 } x + 1 }$$
2. By first using an iterative process based on the equation in part (i), find the coordinates of \(P\), giving each coordinate correct to 3 decimal places.

8. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of part of the curve with equation \(y = \mathrm { f } ( x )\), where \(x \in R\), \(x > 0\) $$\mathrm { f } ( x ) = ( 0.5 x - 8 ) \ln ( x + 1 ) \quad 0 \leq x \leq A$$ a. Find the value of \(A\).
b. Find \(\mathrm { f } ^ { \prime } ( x )\) The curve has a minimum turning point at \(B\).
c. Show that the \(x\)-coordinate of \(B\) is a solution of the equation $$x = \frac { 17 } { \ln ( x + 1 ) + 1 } - 1$$ d. Use the iteration formula $$x _ { n + 1 } = \frac { 17 } { \ln \left( x _ { n } + 1 \right) + 1 } - 1$$ with \(x _ { 0 } = 5\) to find the values of \(x _ { 1 }\) and the value of \(x _ { 6 }\) giving your answers to three decimal places.

6. \begin{figure}[h] \end{figure} Figure 4 shows a sketch of the curve \(C\) with equation \(y = \mathrm { f } ( x )\), where $$f ( x ) = \frac { 2 x ^ { 2 } - x } { \sqrt { x } } - 2 \ln \left( \frac { x } { 2 } \right) , \quad x > 0$$ The curve has a minimum turning point at \(Q\), as shown in Figure 4.
a. Show that \(\mathrm { f } ^ { \prime } ( x ) = \frac { 6 x ^ { 2 } - x - 4 \sqrt { x } } { 2 x \sqrt { x } }\) b. Show that the \(x\)-coordinate of \(Q\) is the solution of $$x = \sqrt { \frac { x } { 6 } + \frac { 2 \sqrt { x } } { 3 } }$$ To find an approximation for the \(x\)-coordinate of \(Q\), the iteration formula $$x _ { n + 1 } = \sqrt { \frac { x _ { n } } { 6 } + \frac { 2 \sqrt { x _ { n } } } { 3 } }$$ is used.
c. Taking \(x _ { 0 } = 0.8\), find the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\). Give your answers to 3 decimal places.

1. The curve with equation \(y = \mathrm { f } ( x )\) where

$$f ( x ) = x ^ { 2 } + \ln \left( 2 x ^ { 2 } - 4 x + 5 \right)$$ has a single turning point at \(x = \alpha\)

1. Show that \(\alpha\) is a solution of the equation $$2 x ^ { 3 } - 4 x ^ { 2 } + 7 x - 2 = 0$$ The iterative formula $$x _ { n + 1 } = \frac { 1 } { 7 } \left( 2 + 4 x _ { n } ^ { 2 } - 2 x _ { n } ^ { 3 } \right)$$ is used to find an approximate value for \(\alpha\).
   Starting with \(x _ { 1 } = 0.3\)
2. calculate, giving each answer to 4 decimal places,
   1. the value of \(x _ { 2 }\)
   2. the value of \(x _ { 4 }\) Using a suitable interval and a suitable function that should be stated,
3. show that \(\alpha\) is 0.341 to 3 decimal places.

6. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of the curve with equation \(y = \mathrm { f } ( x )\), where $$\mathrm { f } ( x ) = ( 8 - x ) \ln x , \quad x > 0$$ The curve cuts the \(x\)-axis at the points \(A\) and \(B\) and has a maximum turning point at \(Q\), as shown in Figure 2.

1. Find the \(x\) coordinate of \(A\) and the \(x\) coordinate of \(B\).
2. Show that the \(x\) coordinate of \(Q\) satisfies $$x = \frac { 8 } { 1 + \ln x }$$
3. Show that the \(x\) coordinate of \(Q\) lies between 3.5 and 3.6
4. Use the iterative formula $$x _ { n + 1 } = \frac { 8 } { 1 + \ln x _ { n } } \quad n \in \mathbb { N }$$ with \(x _ { 1 } = 3.5\) to
   1. find the value of \(x _ { 5 }\) to 4 decimal places,
   2. find the \(x\) coordinate of \(Q\) accurate to 2 decimal places.

11 Fig. 11 shows the graph of \(y = x ^ { 2 } - 4 x + x \ln x\). \begin{figure}[h] \end{figure}

1. Show that the \(x\)-coordinate of the stationary point on the curve may be found from the equation \(2 x - 3 + \ln x = 0\).
2. Use an iterative method to find the \(x\)-coordinate of the stationary point on the curve \(y = x ^ { 2 } - 4 x + x \ln x\), giving your answer correct to 4 decimal places.

4. $$\mathrm { f } ( x ) = 3 \mathrm { e } ^ { x } - \frac { 1 } { 2 } \ln x - 2 , \quad x > 0 .$$

1. Differentiate to find \(\mathrm { f } ^ { \prime } ( x )\). The curve with equation \(y = \mathrm { f } ( x )\) has a turning point at \(P\). The \(x\)-coordinate of \(P\) is \(\alpha\).
2. Show that \(\alpha = \frac { 1 } { 6 } \mathrm { e } ^ { - \alpha }\). The iterative formula $$x _ { n + 1 } = \frac { 1 } { 6 } \mathrm { e } ^ { - x _ { n } } , \quad x _ { 0 } = 1$$ is used to find an approximate value for \(\alpha\).
3. Calculate the values of \(x _ { 1 } , x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving your answers to 4 decimal places.
4. By considering the change of sign of \(\mathrm { f } ^ { \prime } ( x )\) in a suitable interval, prove that \(\alpha = 0.1443\) correct to 4 decimal places.

8. \begin{figure}[h] \end{figure} The heart rate of a horse is being monitored.
The heart rate \(H\), measured in beats per minute (bpm), is modelled by the equation $$H = 32 + 40 \mathrm { e } ^ { - 0.2 t } - 20 \mathrm { e } ^ { - 0.9 t }$$ where \(t\) minutes is the time after monitoring began.
Figure 4 is a sketch of \(H\) against \(t\). \section*{Use the equation of the model to answer parts (a) to (e).}

1. State the initial heart rate of the horse. In the long term, the heart rate of the horse approaches \(L \mathrm { bpm }\).
2. State the value of \(L\). The heart rate of the horse reaches its maximum value after \(T\) minutes.
3. Find the value of \(T\), giving your answer to 3 decimal places.
   (Solutions based entirely on calculator technology are not acceptable.) The heart rate of the horse is 37 bpm after \(M\) minutes.
4. Show that \(M\) is a solution of the equation $$t = 5 \ln \left( \frac { 8 } { 1 + 4 \mathrm { e } ^ { - 0.9 t } } \right)$$ Using the iteration formula $$t _ { n + 1 } = 5 \ln \left( \frac { 8 } { 1 + 4 \mathrm { e } ^ { - 0.9 t _ { n } } } \right) \quad \text { with } \quad t _ { 1 } = 10$$
5. 1. find, to 4 decimal places, the value of \(t _ { 2 }\)
   2. find, to 4 decimal places, the value of \(M\)

The curve \(y = \frac{\ln x}{x + 1}\) has one stationary point.

1. Show that the \(x\)-coordinate of this point satisfies the equation $$x = \frac{x + 1}{\ln x},$$ and that this \(x\)-coordinate lies between 3 and 4. [5]
2. Use the iterative formula $$x_{n+1} = \frac{x_n + 1}{\ln x_n}$$ to determine the \(x\)-coordinate correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

\includegraphics{figure_10} The diagram shows the curve \(y = x^2 \cos 2x\) for \(0 \leq x \leq \frac{1}{4}\pi\). The curve has a maximum point at \(M\) where \(x = p\).

1. Show that \(p\) satisfies the equation \(p = \frac{1}{2} \tan^{-1} \left(\frac{1}{p}\right)\). [3]
2. Use the iterative formula \(p_{n+1} = \frac{1}{2} \tan^{-1} \left(\frac{1}{p_n}\right)\) to determine the value of \(p\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]
3. Find, showing all necessary working, the exact area of the region bounded by the curve and the \(x\)-axis. [5]

The curve with equation \(y = f(x)\) where $$f(x) = x^2 + \ln(2x^2 - 4x + 5)$$ has a single turning point at \(x = \alpha\)

1. Show that \(\alpha\) is a solution of the equation $$2x^3 - 4x^2 + 7x - 2 = 0$$ [4]

The iterative formula $$x_{n+1} = \frac{1}{7}(2 + 4x_n^2 - 2x_n^3)$$ is used to find an approximate value for \(\alpha\). Starting with \(x_1 = 0.3\)

1. calculate, giving each answer to 4 decimal places,
   1. the value of \(x_2\)
   2. the value of \(x_4\)
   [2]

Using a suitable interval and a suitable function that should be stated,

1. show that \(\alpha\) is 0.341 to 3 decimal places. [2]