The problem involves analyzing the function $f(x) = 2x - 1 + \ln(\frac{x}{x+1})$. It asks to find the domain, limits at the boundaries, vertical asymptotes of the graph (C). Next, study the sign of $f'(x)$ knowing that $(x+1) > 0$ for all $x$ in the domain. Calculate extreme values and create a table of variation. Show that the line $(L): y = 2x - 1$ is an oblique asymptote of (C) and find the relative position of (C) and (L). Show the equation $f(x) = 0$ has a unique solution $a \in (0, +\infty)$ and find the approximate solution accurate to $10^{-2}$. Finally, draw the graph (C) and all asymptotes.

1. Problem Description

The problem involves analyzing the function $f(x) = 2x - 1 + \ln(\frac{x}{x+1})$. It asks to find the domain, limits at the boundaries, vertical asymptotes of the graph (C). Next, study the sign of $f'(x)$ knowing that $(x+1) > 0$ for all $x$ in the domain. Calculate extreme values and create a table of variation. Show that the line $(L): y = 2x - 1$ is an oblique asymptote of (C) and find the relative position of (C) and (L). Show the equation $f(x) = 0$ has a unique solution $a \in (0, +\infty)$ and find the approximate solution accurate to $10^{-2}$. Finally, draw the graph (C) and all asymptotes.

2. Solution Steps

a) Determine the domain of definition of $f$.

The function $f(x) = 2x - 1 + \ln(\frac{x}{x+1})$ is defined if $\frac{x}{x+1} > 0$.

This inequality is satisfied if both $x > 0$ and $x + 1 > 0$ or both $x < 0$ and $x+1 < 0$.

If $x > 0$, then $x + 1 > 1 > 0$, so the inequality holds for $x > 0$.

If $x < 0$, we need $x + 1 < 0$, so $x < -1$.

Thus, the domain $D = (-\infty, -1) \cup (0, +\infty)$.

Calculate limits at the boundaries.

$\lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} (2x - 1 + \ln(\frac{x}{x+1})) = \lim_{x \to -\infty} (2x - 1 + \ln(\frac{1}{1+\frac{1}{x}})) = -\infty - 1 + \ln(1) = -\infty$.

$\lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} (2x - 1 + \ln(\frac{x}{x+1})) = 2(-1) - 1 + \lim_{x \to -1^-} \ln(\frac{x}{x+1}) = -3 + \lim_{x \to -1^-} \ln(\frac{-1}{x+1})$.

Since $x \to -1^-$ then $x+1 \to 0^-$, therefore $\frac{-1}{x+1} \to +\infty$, and $\ln(\frac{-1}{x+1}) \to +\infty$.

So $\lim_{x \to -1^-} f(x) = +\infty$. Therefore $x = -1$ is a vertical asymptote.

$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (2x - 1 + \ln(\frac{x}{x+1})) = 0 - 1 + \lim_{x \to 0^+} \ln(\frac{x}{x+1}) = -1 + \lim_{x \to 0^+} \ln(\frac{x}{1}) = -1 - \infty = -\infty$.

So $x = 0$ is a vertical asymptote.

$\lim_{x \to +\infty} f(x) = \lim_{x \to +\infty} (2x - 1 + \ln(\frac{x}{x+1})) = \lim_{x \to +\infty} (2x - 1 + \ln(\frac{1}{1+\frac{1}{x}})) = \lim_{x \to +\infty} (2x - 1 + \ln(1) ) = \lim_{x \to +\infty} (2x - 1) = +\infty$.

b) Study the sign of $f'(x)$.

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

Since $2x^2 + 2x + 1$ has discriminant $\Delta = 4 - 8 = -4 < 0$, so $2x^2 + 2x + 1 > 0$ for all $x$. Thus the sign of $f'(x)$ depends on $x(x+1)$.

If $x \in (-\infty, -1)$, then $x < -1$, so $x < 0$ and $x+1 < 0$, thus $x(x+1) > 0$. Therefore $f'(x) > 0$.

If $x \in (0, +\infty)$, then $x > 0$ and $x+1 > 0$, thus $x(x+1) > 0$. Therefore $f'(x) > 0$.

Thus, $f'(x) > 0$ for all $x \in D$.

$f$ is strictly increasing on $(-\infty, -1)$ and $(0, +\infty)$.

There are no local extrema since the function is increasing.

c) Show that $y = 2x - 1$ is an oblique asymptote of (C).

$\lim_{x \to +\infty} (f(x) - (2x - 1)) = \lim_{x \to +\infty} (2x - 1 + \ln(\frac{x}{x+1}) - (2x - 1)) = \lim_{x \to +\infty} \ln(\frac{x}{x+1}) = \lim_{x \to +\infty} \ln(\frac{1}{1+\frac{1}{x}}) = \ln(1) = 0$.

So $y = 2x - 1$ is an oblique asymptote when $x \to +\infty$.

Relative position of (C) and (L).

Consider $f(x) - (2x - 1) = \ln(\frac{x}{x+1})$.

If $x > 0$, then $\frac{x}{x+1} < 1$, so $\ln(\frac{x}{x+1}) < 0$. Thus, $f(x) < 2x - 1$. Therefore (C) is below (L).

d) Show that $f(x) = 0$ has a unique solution $a \in (0, +\infty)$.

Since $f(x)$ is continuous and strictly increasing on $(0, +\infty)$ and $\lim_{x \to 0^+} f(x) = -\infty$ and $\lim_{x \to +\infty} f(x) = +\infty$, there exists a unique $a \in (0, +\infty)$ such that $f(a) = 0$.

Find the approximate solution to within $10^{-2}$.

$f(1) = 2 - 1 + \ln(\frac{1}{2}) = 1 - \ln(2) \approx 1 - 0.693 = 0.307 > 0$.

$f(0.5) = 2(0.5) - 1 + \ln(\frac{0.5}{1.5}) = 0 + \ln(\frac{1}{3}) = -\ln(3) \approx -1.099 < 0$.

$f(0.75) = 2(0.75) - 1 + \ln(\frac{0.75}{1.75}) = 1.5 - 1 + \ln(\frac{3}{7}) = 0.5 + \ln(3) - \ln(7) \approx 0.5 + 1.099 - 1.946 = -0.347 < 0$.

$f(0.8) = 2(0.8) - 1 + \ln(\frac{0.8}{1.8}) = 1.6 - 1 + \ln(\frac{4}{9}) = 0.6 + 2\ln(2) - 2\ln(3) \approx 0.6 + 2(0.693) - 2(1.099) = 0.6 + 1.386 - 2.198 = -0.212 < 0$.

$f(0.9) = 2(0.9) - 1 + \ln(\frac{0.9}{1.9}) = 1.8 - 1 + \ln(\frac{9}{19}) = 0.8 + \ln(9) - \ln(19) \approx 0.8 + 2.197 - 2.944 = 0.053 > 0$.

$f(0.89) = 2(0.89) - 1 + \ln(\frac{0.89}{1.89}) = 1.78 - 1 + \ln(\frac{89}{189}) \approx 0.78 + \ln(0.4708) \approx 0.78 - 0.7538 = 0.0262 > 0$.

$f(0.88) = 2(0.88) - 1 + \ln(\frac{0.88}{1.88}) = 1.76 - 1 + \ln(\frac{88}{188}) = 0.76 + \ln(0.468) \approx 0.76 - 0.7583 = 0.0017 > 0$

$f(0.87) = 2(0.87) - 1 + \ln(\frac{0.87}{1.87}) = 1.74 - 1 + \ln(\frac{87}{187}) = 0.74 + \ln(0.4652) \approx 0.74 - 0.7642 = -0.0242 < 0$

Therefore $a \approx 0.88$ to the nearest $10^{-2}$.

3. Final Answer

a) Domain: $(-\infty, -1) \cup (0, +\infty)$. Vertical asymptotes: $x = -1$, $x = 0$.

b) $f'(x) > 0$ for all $x$ in the domain.

c) Oblique asymptote: $y = 2x - 1$. (C) is below (L) for $x > 0$.

d) Approximate solution: $a \approx 0.88$.

e) The graph of (C) and the asymptotes can be sketched with the information above.