Problems

Throughout this question, consider only \(x > 0\).

1. Let \[\mathrm{g}(x) = \ln\left(1 + \frac{1}{x}\right) - \frac{x+c}{x(x+1)}\] where \(c \geqslant 0\).
   1. Show that \(y = \mathrm{g}(x)\) has positive gradient for all \(x > 0\) when \(c \geqslant \frac{1}{2}\).
   2. Find the values of \(x\) for which \(y = \mathrm{g}(x)\) has negative gradient when \(0 \leqslant c < \frac{1}{2}\).
2. It is given that, for all \(c > 0\), \(\mathrm{g}(x) \to -\infty\) as \(x \to 0\). Sketch, for \(x > 0\), the graphs of \[y = \mathrm{g}(x)\] in the cases
   1. \(c = \frac{3}{4}\),
   2. \(c = \frac{1}{4}\).
3. The function \(\mathrm{f}\) is defined as \[\mathrm{f}(x) = \left(1 + \frac{1}{x}\right)^{x+c}.\] Show that, for \(x > 0\),
   1. \(\mathrm{f}\) is a decreasing function when \(c \geqslant \frac{1}{2}\);
   2. \(\mathrm{f}\) has a turning point when \(0 < c < \frac{1}{2}\);
   3. \(\mathrm{f}\) is an increasing function when \(c = 0\).