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 \(x = \alpha\).
Show that \(\alpha\) lies between 0 and \(\frac { \pi } { 2 }\).
Show that the equation \(\frac { \cos x } { 2 x + 1 } = \frac { 1 } { 2 }\) can be rearranged into the form
$$x = \cos x - \frac { 1 } { 2 }$$
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.
Given that \(y = \frac { \cos x } { 2 x + 1 }\), use the quotient rule to find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
Hence find the gradient of the normal to the curve \(y = \frac { \cos x } { 2 x + 1 }\) at the point on the curve where \(x = 0\).