The problem provides a function $f(x) = -x + 4 + \ln(\frac{x+1}{x-1})$ defined on the interval $(1, +\infty)$. The questions ask to compute the limits of $f$ at $1$ and $+\infty$, to study the variation of $f$, to prove that the line $y = -x + 4$ is an asymptote to the graph of $f$ as $x \to +\infty$, to determine the relative position of the graph of $f$ with respect to the line $y = -x+4$, to find the point where the tangent line to the curve is parallel to a line with slope $-\frac{5}{3}$, and to plot the curves and lines.

1. Problem Description

The problem provides a function $f(x) = -x + 4 + \ln(\frac{x+1}{x-1})$ defined on the interval $(1, +\infty)$. The questions ask to compute the limits of $f$ at $1$ and $+\infty$, to study the variation of $f$, to prove that the line $y = -x + 4$ is an asymptote to the graph of $f$ as $x \to +\infty$, to determine the relative position of the graph of $f$ with respect to the line $y = -x+4$, to find the point where the tangent line to the curve is parallel to a line with slope $-\frac{5}{3}$, and to plot the curves and lines.

2. Solution Steps

a) Calculate the limits of $f$ at $1$ and $+\infty$.

$\lim_{x \to 1^{+}} f(x) = \lim_{x \to 1^{+}} \left(-x + 4 + \ln\left(\frac{x+1}{x-1}\right)\right) = -1 + 4 + \lim_{x \to 1^{+}} \ln\left(\frac{x+1}{x-1}\right)$.

As $x \to 1^{+}$, $x-1 \to 0^{+}$, so $\frac{x+1}{x-1} \to \frac{2}{0^{+}} = +\infty$. Thus, $\ln\left(\frac{x+1}{x-1}\right) \to +\infty$.

Therefore, $\lim_{x \to 1^{+}} f(x) = 3 + \infty = +\infty$.

$\lim_{x \to +\infty} f(x) = \lim_{x \to +\infty} \left(-x + 4 + \ln\left(\frac{x+1}{x-1}\right)\right)$.

We have $\lim_{x \to +\infty} \frac{x+1}{x-1} = \lim_{x \to +\infty} \frac{1 + \frac{1}{x}}{1 - \frac{1}{x}} = \frac{1+0}{1-0} = 1$. So $\lim_{x \to +\infty} \ln\left(\frac{x+1}{x-1}\right) = \ln(1) = 0$.

Therefore, $\lim_{x \to +\infty} f(x) = \lim_{x \to +\infty} (-x + 4) + 0 = -\infty$.

b) Show that $f'(x) = -\frac{x^2+1}{(x-1)(x+1)}$. Study the variations of $f$.

$f(x) = -x + 4 + \ln(x+1) - \ln(x-1)$.

$f'(x) = -1 + \frac{1}{x+1} - \frac{1}{x-1} = -1 + \frac{x-1 - (x+1)}{(x+1)(x-1)} = -1 + \frac{-2}{x^2-1} = \frac{-x^2 + 1 - 2}{x^2 - 1} = \frac{-x^2 - 1}{x^2 - 1} = -\frac{x^2+1}{x^2-1} = -\frac{x^2+1}{(x-1)(x+1)}$.

Since $x > 1$, $x^2 + 1 > 0$ and $(x-1)(x+1) > 0$, so $f'(x) < 0$.

Therefore, $f$ is decreasing on $(1, +\infty)$.

c) Show that the line $y = -x + 4$ is an asymptote to the graph of $f$ as $x \to +\infty$.

Consider $f(x) - (-x + 4) = -x + 4 + \ln(\frac{x+1}{x-1}) - (-x + 4) = \ln(\frac{x+1}{x-1})$.

We have $\lim_{x \to +\infty} \ln(\frac{x+1}{x-1}) = \ln(1) = 0$.

Therefore, the line $y = -x + 4$ is an asymptote to the graph of $f$ as $x \to +\infty$.

Show that $\frac{x+1}{x-1} > 1$ and deduce the relative position of the graph of $f$ with respect to the line $y = -x+4$.

Since $x > 1$, $x - 1 > 0$, so $\frac{x+1}{x-1} > 1$ if and only if $x+1 > x-1$, which is equivalent to $1 > -1$. This is always true for $x > 1$.

Then, $\ln(\frac{x+1}{x-1}) > \ln(1) = 0$.

Therefore, $f(x) - (-x + 4) = \ln(\frac{x+1}{x-1}) > 0$, which implies $f(x) > -x + 4$.

The graph of $f$ is above the line $y = -x + 4$.

d) Find the point where the tangent line to the curve has slope $-\frac{5}{3}$.

We need to find $x$ such that $f'(x) = -\frac{5}{3}$.

$-\frac{x^2+1}{x^2-1} = -\frac{5}{3}$

$3(x^2+1) = 5(x^2-1)$

$3x^2 + 3 = 5x^2 - 5$

$2x^2 = 8$

$x^2 = 4$

Since $x > 1$, $x = 2$.

$f(2) = -2 + 4 + \ln(\frac{2+1}{2-1}) = 2 + \ln(3)$.

So the point is $(2, 2+\ln(3))$. The tangent line has the form $y = -\frac{5}{3}(x-2) + 2 + \ln(3)$.

$y = -\frac{5}{3}x + \frac{10}{3} + 2 + \ln(3) = -\frac{5}{3}x + \frac{16}{3} + \ln(3)$.

Therefore, the equation of the tangent line $(d_2)$ is $y = -\frac{5}{3}x + \frac{16}{3} + \ln(3)$.

3. Final Answer

a) $\lim_{x \to 1^{+}} f(x) = +\infty$, $\lim_{x \to +\infty} f(x) = -\infty$.

b) $f'(x) = -\frac{x^2+1}{(x-1)(x+1)}$. $f$ is decreasing on $(1, +\infty)$.

c) $y = -x + 4$ is an asymptote. The graph of $f$ is above the line $y = -x + 4$.

d) The point is $(2, 2+\ln(3))$. The equation of the tangent line $(d_2)$ is $y = -\frac{5}{3}x + \frac{16}{3} + \ln(3)$.