Derive equation from calculus condition

7 The curve with equation $$y = \frac { x } { \cosh x }$$ has one stationary point for \(x > 0\).

1. Show that the \(x\)-coordinate of this stationary point satisfies the equation \(x \tanh x - 1 = 0\). The positive root of the equation \(x \tanh x - 1 = 0\) is denoted by \(\alpha\).
2. Draw a sketch showing (for positive values of \(x\) ) the graph of \(y = \tanh x\) and its asymptote, and the graph of \(y = \frac { 1 } { x }\). Explain how you can deduce from your sketch that \(\alpha > 1\).
3. Use the Newton-Raphson method, taking first approximation \(x _ { 1 } = 1\), to find further approximations \(x _ { 2 }\) and \(x _ { 3 }\) for \(\alpha\).
4. By considering the approximate errors in \(x _ { 1 }\) and \(x _ { 2 }\), estimate the error in \(x _ { 3 }\).

8. $$f ( x ) = 2 x + \sin x - 3 \cos x$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root in the interval [0.7, 0.8].
2. Find an equation for the tangent to the curve \(y = \mathrm { f } ( x )\) at the point where it crosses the \(y\)-axis.
3. Find the values of the constants \(a , b\) and \(c\), where \(b > 0\) and \(0 < c < \frac { \pi } { 2 }\), such that $$f ^ { \prime } ( x ) = a + b \cos ( x - c )$$
4. Hence find the \(x\)-coordinates of the stationary points of the curve \(y = \mathrm { f } ( x )\) in the interval \(0 \leq x \leq 2 \pi\), giving your answers to 2 decimal places.

7 [Figure 1, printed on the insert, is provided for use in this question.]
The diagram shows the curve $$y = x ^ { 3 } - x + 1$$ The points \(A\) and \(B\) on the curve have \(x\)-coordinates - 1 and \(- 1 + h\) respectively. \includegraphics[max width=\textwidth, alt={}, center]{a0a30197-ca11-40d9-9ccd-30281c5e0fb4-05_978_1184_676_411}

1. 1. Show that the \(y\)-coordinate of the point \(B\) is $$1 + 2 h - 3 h ^ { 2 } + h ^ { 3 }$$
   2. Find the gradient of the chord \(A B\) in the form $$p + q h + r h ^ { 2 }$$ where \(p , q\) and \(r\) are integers.
   3. Explain how your answer to part (a)(ii) can be used to find the gradient of the tangent to the curve at \(A\). State the value of this gradient.
2. The equation \(x ^ { 3 } - x + 1 = 0\) has one real root, \(\alpha\).
   1. Taking \(x _ { 1 } = - 1\) as a first approximation to \(\alpha\), use the Newton-Raphson method to find a second approximation, \(x _ { 2 }\), to \(\alpha\).
   2. On Figure 1, draw a straight line to illustrate the Newton-Raphson method as used in part (b)(i). Show the points \(\left( x _ { 2 } , 0 \right)\) and \(( \alpha , 0 )\) on your diagram.

In this question you must show detailed reasoning. \includegraphics{figure_5} The diagram shows the curve with equation \(y = \frac{2x - 3}{4x^2 + 1}\). The tangent to the curve at the point \(P\) has gradient 2.

1. Show that the \(x\)-coordinate of \(P\) satisfies the equation $$4x^3 + 3x - 3 = 0.$$ [5]
2. Show by calculation that the \(x\)-coordinate of \(P\) lies between 0.5 and 1. [2]
3. Show that the iteration $$x_{n+1} = \frac{3 - 4x_n^3}{3}$$ cannot converge to the \(x\)-coordinate of \(P\) whatever starting value is used. [2]
4. Use the Newton-Raphson method, with initial value 0.5, to determine the coordinates of \(P\) correct to 5 decimal places. [5]