We are given the function $g(x) = x^2 - 1 + \ln x$ defined on the interval $]0, +\infty[$. We need to find its derivative $g'(x)$, deduce the sense of variation of $g$, calculate $g(1)$, and deduce the sign of $g(x)$.

1. Problem Description

We are given the function $g(x) = x^2 - 1 + \ln x$ defined on the interval $]0, +\infty[$. We need to find its derivative $g'(x)$, deduce the sense of variation of $g$, calculate $g(1)$, and deduce the sign of $g(x)$.

2. Solution Steps

1. Calculating $g'(x)$:

We have $g(x) = x^2 - 1 + \ln x$. We need to find the derivative $g'(x)$.

The derivative of $x^2$ is $2x$.

The derivative of $-1$ is $0$.

The derivative of $\ln x$ is $\frac{1}{x}$.

Therefore,

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

2. Deduce the sense of variation of $g$ on $]0, +\infty[$:

Since $x > 0$ on the interval $]0, +\infty[$, we have $2x > 0$ and $\frac{1}{x} > 0$. Thus, $g'(x) = 2x + \frac{1}{x} > 0$ for all $x \in ]0, +\infty[$.

This means that the function $g(x)$ is strictly increasing on the interval $]0, +\infty[$.

3. Calculating $g(1)$:

$g(1) = (1)^2 - 1 + \ln(1) = 1 - 1 + 0 = 0$.

4. Deduce the sign of $g(x)$ for $x \in ]0, +\infty[$:

Since $g(x)$ is strictly increasing on $]0, +\infty[$ and $g(1) = 0$, we can analyze the sign of $g(x)$ based on the value of $x$.

If $0 < x < 1$, then $g(x) < g(1) = 0$.

If $x > 1$, then $g(x) > g(1) = 0$.

If $x = 1$, then $g(x) = g(1) = 0$.

So,

$g(x) < 0$ for $x \in ]0, 1[$,

$g(x) = 0$ for $x = 1$,

$g(x) > 0$ for $x \in ]1, +\infty[$.

3. Final Answer

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

The function $g(x)$ is strictly increasing on the interval $]0, +\infty[$.

2. $g(1) = 0$

$g(x) < 0$ for $x \in ]0, 1[$,

$g(x) = 0$ for $x = 1$,

$g(x) > 0$ for $x \in ]1, +\infty[$.