We are given the function $f(x) = 2x - 1 + \ln(\frac{x}{x+1})$. The problem asks us to find the domain of $f$, calculate limits at the boundaries of the domain, find the vertical asymptotes, study the sign of $f'(x)$ knowing that $\ln(x+1)>0$, calculate the extreme values and create the table of variations. Further, show that the line $y = 2x - 1$ is an oblique asymptote of $(C)$ and determine the relative position between $(C)$ and $(L)$. Finally, prove that the equation $f(x) = 0$ has a unique solution $a \in (0, +\infty)$, approximate the solution to $10^{-1}$, and sketch the graph of $(C)$ and all asymptotes.

1. Problem Description

We are given the function $f(x) = 2x - 1 + \ln(\frac{x}{x+1})$. The problem asks us to find the domain of $f$, calculate limits at the boundaries of the domain, find the vertical asymptotes, study the sign of $f'(x)$ knowing that $\ln(x+1)>0$, calculate the extreme values and create the table of variations. Further, show that the line $y = 2x - 1$ is an oblique asymptote of $(C)$ and determine the relative position between $(C)$ and $(L)$. Finally, prove that the equation $f(x) = 0$ has a unique solution $a \in (0, +\infty)$, approximate the solution to $10^{-1}$, and sketch the graph of $(C)$ and all asymptotes.

2. Solution Steps

(a) Domain of $f(x)$:

For $f(x)$ to be defined, we need $\frac{x}{x+1} > 0$.

This inequality holds if both $x$ and $x+1$ are positive or both $x$ and $x+1$ are negative.

Case 1: $x > 0$ and $x+1 > 0$. This gives $x > 0$ and $x > -1$. Thus, $x > 0$.

Case 2: $x < 0$ and $x+1 < 0$. This gives $x < 0$ and $x < -1$. Thus, $x < -1$.

Therefore, the domain of $f(x)$ is $D = (-\infty, -1) \cup (0, +\infty)$.

Limits and vertical asymptotes:

We calculate the limits at the boundaries of the domain.

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

$\lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} (2x - 1 + \ln(\frac{x}{x+1})) = 2(-1) - 1 + \ln(\frac{-1}{-1^-+1}) = -3 + \ln(\frac{-1}{0^-}) = -3 + \ln(+\infty) = +\infty$. Thus, $x = -1$ is a vertical asymptote.

$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (2x - 1 + \ln(\frac{x}{x+1})) = 2(0) - 1 + \ln(\frac{0^+}{0^++1}) = -1 + \ln(0^+) = -1 - \infty = -\infty$. Thus, $x = 0$ is a vertical asymptote.

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

(b) Sign of $f'(x)$:

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

The numerator $2x^2 + 2x + 1$ has discriminant $\Delta = 2^2 - 4(2)(1) = 4 - 8 = -4 < 0$. Therefore, $2x^2 + 2x + 1 > 0$ for all $x$.

The sign of $f'(x)$ depends on the sign of $x(x+1)$.

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

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

Therefore, $f'(x) > 0$ for all $x$ in the domain of $f(x)$. This means that $f(x)$ is strictly increasing on $(-\infty, -1)$ and $(0, +\infty)$.

Since $f'(x) > 0$, there are no extreme values.

Table of variations:

$x$ | $-\infty$ | | $-1$ | | $0$ | | $+\infty$

----|----------|-----|-------|-----|------|-----|----------

$f'(x)$| | $+$ | | $+$ | | $+$ |

$f(x)$ | $-\infty$ | $\nearrow$ | $+\infty$ | $\nearrow$ | $-\infty$ | $\nearrow$ | $+\infty$

(c) Oblique Asymptote:

We need to show that $\lim_{x \to \pm \infty} [f(x) - (2x - 1)] = 0$.

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

Therefore, $y = 2x - 1$ is an oblique asymptote.

Relative position:

We study the sign of $f(x) - (2x - 1) = \ln(\frac{x}{x+1})$.

Since $\frac{x}{x+1} > 0$ on the domain,

$f(x) - (2x-1) > 0$ if $\frac{x}{x+1} > 1 \implies \frac{x}{x+1} - 1 > 0 \implies \frac{x - (x+1)}{x+1} > 0 \implies \frac{-1}{x+1} > 0 \implies x+1 < 0 \implies x < -1$.

$f(x) - (2x-1) < 0$ if $\frac{x}{x+1} < 1 \implies \frac{-1}{x+1} < 0 \implies x+1 > 0 \implies x > -1$. Since $x > 0$, $f(x) < 2x-1$.

(d) Solution of $f(x) = 0$:

On $(-\infty, -1)$, $f(x)$ is strictly increasing from $-\infty$ to $+\infty$. Thus, there exists a unique solution. Since we are only concerned with the interval $(0, +\infty)$, we can ignore this.

On $(0, +\infty)$, $f(x)$ is strictly increasing from $-\infty$ to $+\infty$. Thus, there exists a unique solution $a \in (0, +\infty)$ to $f(x) = 0$.

Let's approximate the solution. We want to find an $x$ such that $f(x) \approx 0$.

$f(0.1) = 2(0.1) - 1 + \ln(\frac{0.1}{1.1}) = 0.2 - 1 + \ln(\frac{1}{11}) \approx -0.8 - 2.4 = -3.2$

$f(1) = 2(1) - 1 + \ln(\frac{1}{2}) = 1 + \ln(0.5) = 1 - 0.693 = 0.307$

Since $f(0.1) < 0$ and $f(1) > 0$, the solution is between $0.1$ and $1$.

$f(0.2) = 2(0.2) - 1 + \ln(\frac{0.2}{1.2}) = 0.4 - 1 + \ln(\frac{1}{6}) = -0.6 + \ln(0.167) \approx -0.6 - 1.79 = -2.39$

$f(0.5) = 2(0.5) - 1 + \ln(\frac{0.5}{1.5}) = 1 - 1 + \ln(\frac{1}{3}) = \ln(0.333) = -1.09$

Since $f(1) > 0$ and $f(0.5) < 0$, the root is between 0.5 and

1. $f(0.8) = 2(0.8) - 1 + \ln(\frac{0.8}{1.8}) = 1.6 - 1 + \ln(\frac{8}{18}) = 0.6 + \ln(\frac{4}{9}) \approx 0.6 + \ln(0.444) \approx 0.6 - 0.81 = -0.21$

Since $f(1) > 0$ and $f(0.8) < 0$, the root is between 0.8 and

1.

$f(0.9) = 2(0.9) - 1 + \ln(\frac{0.9}{1.9}) = 1.8 - 1 + \ln(0.474) \approx 0.8 - 0.747 = 0.053 > 0$

So the root is between $0.8$ and $0.9$. We want an approximation to $10^{-1} = 0.1$. Let's check if the root is closer to $0.8$ or $0.9$.

Since $|f(0.9)| \approx 0.053$ and $|f(0.8)| \approx 0.21$, it is closer to $0.9$.

Therefore, to the nearest tenth, the solution is approximately $0.9$.

3. Final Answer

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

(b) $f'(x) > 0$ on the domain, so the function is strictly increasing. There are no extreme values.

(c) $y = 2x - 1$ is an oblique asymptote. If $x < -1$, $f(x) > 2x - 1$. If $x > 0$, $f(x) < 2x - 1$.

(d) The equation $f(x) = 0$ has a unique solution in $(0, +\infty)$. An approximate solution is $0.9$.