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Notice that by partial fractions, we have \[ \frac {x + c}{x (x + 1)} = \frac {1 - c}{x + 1} + \frac {c}{x}. \]
Hence, by differentiating, we have \begin {align*} g'(x) & = \frac {1}{1 + \frac {1}{x}} \cdot \left (- \frac {1}{x^2}\right ) + \frac {1 - c}{(x + 1)^2} + \frac {c}{x^2} \\ & = - \frac {1}{x^2 + x} + \frac {1 - c}{(x + 1)^2} + \frac {c}{x^2} \\ & = \frac {- x (x + 1) + (1 - c) x^2 + c (x + 1)^2}{(x + 1)^2 x^2} \\ & = \frac {cx^2 + 2cx + c + x^2 - cx^2 - x^2 - x}{(x + 1)^2 x^2} \\ & = \frac {(2c - 1)x + c}{(x + 1)^2 x^2}. \end {align*}
Given that \(c \geq \frac {1}{2}\), and \(x > 0\), we have \(2c - 1 \geq 0\), and \((2c - 1) x \geq 0\).
Hence, the numerator satisfies \((2c - 1)x + c \geq c \geq \frac {1}{2} > 0\), and the denominator is always positive since is a product of squares, and both squares are non-zero since \(x > 0\).
We can now conclude that \(g'(x) > 0\) given \(c \geq \frac {1}{2}\) for \(x > 0\), as desired.
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If \(0 \leq c < \frac {1}{2}\), \(g'(x) < 0\) if and only if \begin {align*} (2c - 1) x + c & < 0 \\ (1 - 2c) x - c & > 0 \\ (1 - 2c) x & > c \\ x & > \frac {c}{1 - 2c}, \end {align*}
and the values of \(x\) are \(x > \frac {c}{1 - 2c}\).
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If \(c = \frac {3}{4} \geq \frac {1}{2}\), we can see that \(g\) is always increasing.
As \(x \to \infty \), \(\frac {x + c}{x (x + 1)} \to 0\), \(\ln \left (1 + \frac {1}{x}\right ) \to \ln 1 = 0\). Hence, \(g(x) \to 0\).
Since \(g\) is increasing it must stay entirely below the \(x\)-axis.
The sketch is as follows.
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If \(c = \frac {1}{4} \in \left [0, \frac {1}{2}\right )\), it must be the case that \(g'(x) > 0\) for \(0 < x < \frac {c}{1 - 2c} = \frac {1}{2}\), and \(g'(x) < 0\) for \(x > \frac {1}{2}\).
Hence, \(x = \frac {1}{2}\) is a maximum on the graph, and the corresponding \(y\)-coordinate is \(g\left (\frac {1}{2}\right ) = \ln 3 - 1\).
Similarly, as \(x \to \infty \), \(g(x) \to 0\).
The sketch is as follows.
We have \begin {align*} f(x) & = \left (1 + \frac {1}{x}\right )^{x + c} \\ \ln f(x) & = (x + c) \ln \left (1 + \frac {1}{x}\right ) \\ \frac {f'(x)}{f(x)} & = \ln \left (1 + x\right ) - (x + c) \frac {1}{x (x + 1)} \\ \frac {f'(x)}{f(x)} & = g(x) \\ f'(x) & = f(x) g(x). \end {align*}
\(f(x)\) is positive for \(x > 0\), and hence \(f'(x)\) takes the same sign as \(g(x)\).
- If \(c \geq \frac {1}{2}\), \(g\) is increasing and has a limit of \(0\) at infinity. Hence, \(g(x)\) is negative for all \(x > 0\), which means \(f'(x)\) is negative for all \(x > 0\), and hence \(f\) is decreasing.
- If \(0 < c < \frac {1}{2}\), \(g\) is negative first, then increases to a positive value, and remains positive and approaches \(0\) decreasing from above. Hence, \(f'\) is first positive and then negative, so \(f\) must have a turning point.
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If \(c = 0\), \[ g'(x) = \frac {-x}{(x + 1)^2 x^2} = - \frac {1}{(x + 1)^2 x} \] is always negative, and \(\lim _{x \to 0^{+}} g'(x) =-\infty \), \(\lim _{x \to \infty } g'(x) = 0\).
We have \[ g(x) = \ln \left (1 + \frac {1}{x}\right ) - \frac {1}{x + 1}. \]
As \(x \to 0^{+}\), \(\frac {1}{x} \to \infty \), so \(\ln \left (1 + \frac {1}{x}\right ) \to \infty \), and \(- \frac {1}{x + 1} \to - \frac {1}{1} = -1\). Hence, \(g(x) \to \infty \).
As \(x \to \infty \), \(g(x) \to 0\).
Since \(g\) is decreasing, it must be the case that \(g\) is always positive.
This means that \(f'\) is always positive as well, and hence \(f\) is increasing.