We are given two functions, $g(x) = -\frac{1}{x} + \ln x$ defined on $(0, +\infty)$ and $f(x) = x - (x-1)\ln x$ also defined on $(0, +\infty)$. We are asked to find various properties of these functions, including limits, derivatives, variations, and graphical representation.

1. Problem Description

We are given two functions, $g(x) = -\frac{1}{x} + \ln x$ defined on $(0, +\infty)$ and $f(x) = x - (x-1)\ln x$ also defined on $(0, +\infty)$. We are asked to find various properties of these functions, including limits, derivatives, variations, and graphical representation.

2. Solution Steps

Partie A:

1. Calculate the limits of $g$ at $0^+$ and $+\infty$.

As $x \to 0^+$, $-\frac{1}{x} \to -\infty$ and $\ln x \to -\infty$. Therefore,

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

As $x \to +\infty$, $-\frac{1}{x} \to 0$ and $\ln x \to +\infty$. Therefore,

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

2. a. Calculate the derivative $g'$ of $g$ on $(0, +\infty)$.

$g'(x) = \frac{d}{dx} \left( -\frac{1}{x} + \ln x \right) = \frac{d}{dx} (-x^{-1}) + \frac{d}{dx} (\ln x) = x^{-2} + \frac{1}{x} = \frac{1}{x^2} + \frac{1}{x} = \frac{1+x}{x^2}$.

b. Dress the variation table of $g$.

Since $x>0$, we have $x^2 > 0$ and $1+x > 0$, so $g'(x) = \frac{1+x}{x^2} > 0$. This means $g(x)$ is strictly increasing on $(0, +\infty)$. We know $\lim_{x \to 0^+} g(x) = -\infty$ and $\lim_{x \to +\infty} g(x) = +\infty$.

3. a. Show that the equation $g(x) = 0$ admits a unique solution $\alpha \in [\frac{3}{2}, 2]$.

Since $g(x)$ is continuous and strictly increasing on $(0, +\infty)$, it admits at most one solution to $g(x) = 0$. Since $\lim_{x \to 0^+} g(x) = -\infty$ and $\lim_{x \to +\infty} g(x) = +\infty$, then by the intermediate value theorem, there is a solution to $g(x)=0$. Thus, the equation $g(x)=0$ admits a unique solution.

We check the sign of $g(x)$ at $x=\frac{3}{2}$ and $x=2$.

$g\left(\frac{3}{2}\right) = -\frac{2}{3} + \ln\left(\frac{3}{2}\right) \approx -0.667 + 0.405 = -0.262 < 0$.

$g(2) = -\frac{1}{2} + \ln(2) \approx -0.5 + 0.693 = 0.193 > 0$.

Since $g(\frac{3}{2}) < 0$ and $g(2) > 0$, and $g$ is continuous, there exists a value $\alpha \in [\frac{3}{2}, 2]$ such that $g(\alpha) = 0$. Since $g$ is strictly increasing, this solution is unique.

b. Deduce the sign of $g(x)$ on $(0, +\infty)$.

Since $g(x)$ is strictly increasing and $g(\alpha) = 0$, we have:

$g(x) < 0$ for $x \in (0, \alpha)$

$g(x) = 0$ for $x = \alpha$

$g(x) > 0$ for $x \in (\alpha, +\infty)$

Partie B:

1. Calculate the limits of $f$ at $0^+$ and $+\infty$.

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

As $x \to 0^+$, $x \to 0$ and $\ln x \to -\infty$.

$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} [x - (x-1)\ln x] = 0 - \lim_{x \to 0^+} [x\ln x - \ln x]$.

We know $\lim_{x \to 0^+} x\ln x = 0$. Thus

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

As $x \to +\infty$, $x \to +\infty$ and $\ln x \to +\infty$.

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

Since $\ln x$ grows slower than $x$, for sufficiently large $x$, $\ln x > 1$, so $1-\ln x < 0$. Thus,

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

However, $x$ grows faster than $\ln x$. Thus the limit is $-\infty$.

$\lim_{x \to +\infty} f(x) = -\infty$.

2. a. Show that the derivative $f'$ of $f$ is $f'(x) = -g(x)$.

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

$f'(x) = 1 - (\ln x + x \cdot \frac{1}{x}) + \frac{1}{x} = 1 - \ln x - 1 + \frac{1}{x} = \frac{1}{x} - \ln x = - \left( -\frac{1}{x} + \ln x \right) = -g(x)$.

b. Draw the variation table of $f$.

We know that $f'(x) = -g(x)$.

Since $g(x) < 0$ for $x \in (0, \alpha)$ and $g(x) > 0$ for $x \in (\alpha, +\infty)$, then $f'(x) > 0$ for $x \in (0, \alpha)$ and $f'(x) < 0$ for $x \in (\alpha, +\infty)$. This implies that $f$ is increasing on $(0, \alpha)$ and decreasing on $(\alpha, +\infty)$. Also $f'(\alpha) = 0$.

Given that $\alpha \approx 1.7$ and $f(\alpha) \approx 1.3$.

3. On admet que l'équation $f(x)=0$, admet deux solutions $x_0$ et $x_1$ avec $x_0 \in [\frac{1}{2}, 1]$ et $x_1 \in [\frac{7}{2}, 4]$.

a. Étudier la branche infinie à (C). Replace the question with: « Étudier les branches infinies (C). »

We are asked to find the limit $\lim_{x \to +\infty} \frac{f(x)}{x}$.

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

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

Since the limit is $-\infty$, there is a parabolic branch in the direction of the y-axis.

b. Trace the curve (C).

The curve (C) can be sketched with the information gathered.

4. Trace the curve (C') of the function $h$ defined by $h(x)=-f(x)$ in the same coordinate system as (C).

Since $h(x) = -f(x)$, the curve (C') is a reflection of (C) across the x-axis.

3. Final Answer

Partie A:

1. $\lim_{x \to 0^+} g(x) = -\infty$, $\lim_{x \to +\infty} g(x) = +\infty$

2. a. $g'(x) = \frac{1+x}{x^2}$

b. $g(x)$ is strictly increasing on $(0, +\infty)$

3. a. $\alpha \in [\frac{3}{2}, 2]$ is the unique solution to $g(x) = 0$

b. $g(x) < 0$ for $x \in (0, \alpha)$, $g(x) = 0$ for $x = \alpha$, $g(x) > 0$ for $x \in (\alpha, +\infty)$

Partie B:

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

2. a. $f'(x) = -g(x)$

3. a. There is a parabolic branch in the direction of the y-axis.