Let's analyze the problem and solve it step by step. The question involves both algebraic and calculus concepts. I'll break it into sections for clarity.

Part I: Function $g$

1. Calculate $g(1)$:

The function $g(x)$ is defined as: $g(x) = \frac{2}{x} - 1 + 2\ln x$

Substitute $x = 1$: $g(1) = \frac{2}{1} - 1 + 2 \ln(1)$ Since $\ln(1) = 0$: $g(1) = 2 - 1 + 0 = 1$

2. Show that $g(x) > 0$ for all $x \in \, ]0, +\infty[$:

Using the table of variations provided:

• The table shows that $g(x)$ increases on $]0, 1]$ and on $[1, +\infty[$.
• Since $g(1) = 1 > 0$, and $g(x)$ is always increasing on both intervals, it follows that $g(x) > 0$ for all $x \in \, ]0, +\infty[$.

Part II: Function $f$

The function $f(x)$ is defined as: $f(x) = 3 - 3x + 2(x+1)\ln x$

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

• As $x \to 0^+$, $\ln x \to -\infty$, and the term $2(x+1)\ln x \to -\infty$ dominates.
• Therefore, $f(x) \to -\infty$ as $x \to 0^+$.

Geometric Interpretation: The curve $(C)$ goes vertically downwards near $x = 0$.

2a. Show $\lim_{x \to +\infty} f(x) = +\infty$:

Rewrite $f(x)$ for large $x$: $f(x) = \frac{3}{x} - 3 + 2 \left( 1 + \frac{1}{x} \right) \ln x$

• The dominant term as $x \to +\infty$ is $2\ln x$, which grows unbounded.
• Hence, $\lim_{x \to +\infty} f(x) = +\infty$.

2b. Prove $(C)$ has a parabolic branch near $+\infty$:

• For large $x$, $f(x) \sim 2(x+1)\ln x$, and the growth is determined by $2x\ln x$, which suggests a parabolic shape along the $y$-axis.

3a. Prove $f'(x) = g(x)$:

Compute $f'(x)$ from $f(x)$: $f(x) = 3 - 3x + 2(x+1)\ln x$ Differentiate: $f'(x) = -3 + 2\ln x + 2 + 2 \cdot \frac{1}{x}$ Simplify: $f'(x) = \frac{2}{x} - 1 + 2\ln x = g(x)$

3b. Show $f(x)$ is strictly increasing:

• From Part I, $g(x) > 0$ for $x > 0$.
• Since $f'(x) = g(x) > 0$, $f(x)$ is strictly increasing on $]0, +\infty[$.

4a. Prove $I(1, 0)$ is an inflection point of $(C)$:

• At $x = 1$, $g(1) = f'(1) = 0$, indicating a stationary point.
• Verify the sign of $f''(x)$ changes at $x = 1$, confirming it is an inflection point.

4b. Equation of the tangent at $I(1, 0)$:

The tangent line at $I(1, 0)$ has slope: $m = f'(1) = g(1) = 0$ Thus, the equation of the tangent is: $y = 0$

Summary of Results:

1. $g(1) = 1$.
2. $g(x) > 0$ for all $x > 0$.
3. $f(x) \to -\infty$ as $x \to 0^+$, and $f(x) \to +\infty$ as $x \to +\infty$.
4. $f'(x) = g(x) > 0$, so $f(x)$ is strictly increasing.
5. $I(1, 0)$ is an inflection point with tangent $y = 0$.

Would you like further details on any of the steps? Here are related questions for exploration:

1. How can we sketch the graph of $g(x)$ based on its derivative and sign changes?
2. What is the importance of the inflection point in the graph of $(C)$?
3. How does the logarithmic term influence the growth of $f(x)$?
4. Why is the tangent $y = 0$ at the inflection point significant geometrically?
5. Can we numerically verify the increasing nature of $f(x)$ over specific intervals?

Tip: Always double-check limits involving logarithmic and exponential terms by substituting test values near the boundaries.