We are given the function $f(x) = \frac{x}{x^2 + 1}$, and we need to determine the largest possible value of $b - a$ where the function is increasing on the interval $]a, b[$.

Step 1: Find the derivative of the function $f(x)$

The derivative of a function tells us whether the function is increasing or decreasing in a given interval. So, we will calculate the derivative of $f(x) = \frac{x}{x^2 + 1}$ using the quotient rule.

The quotient rule states that for $f(x) = \frac{u(x)}{v(x)}$, the derivative is given by: $f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$ Here, $u(x) = x$ and $v(x) = x^2 + 1$, so:

• $u'(x) = 1$
• $v'(x) = 2x$

Now applying the quotient rule: $f'(x) = \frac{(1)(x^2 + 1) - (x)(2x)}{(x^2 + 1)^2}$ Simplifying the numerator: $f'(x) = \frac{x^2 + 1 - 2x^2}{(x^2 + 1)^2} = \frac{-x^2 + 1}{(x^2 + 1)^2}$ $f'(x) = \frac{1 - x^2}{(x^2 + 1)^2}$

Step 2: Analyze where $f'(x)$ is positive

For the function to be increasing, the derivative must be positive, i.e., $f'(x) > 0$. Therefore, we need: $\frac{1 - x^2}{(x^2 + 1)^2} > 0$ The denominator $(x^2 + 1)^2$ is always positive, so the sign of $f'(x)$ depends on the numerator, $1 - x^2$.

Thus, the inequality becomes: $1 - x^2 > 0$ $x^2 < 1$ This implies: $-1 < x < 1$

Step 3: Conclusion

The function is increasing on the interval $]-1, 1[$. Therefore, $a = -1$ and $b = 1$, so: $b - a = 1 - (-1) = 2$

Thus, the greatest value of $b - a$ is $\boxed{2}$, which corresponds to option (c).

Let me know if you need further details or explanations!

Here are 5 related questions for deeper understanding:

1. How does the quotient rule work for derivatives of rational functions?
2. Why is the denominator of $f'(x)$ always positive in this case?
3. Can we have an interval where the function is decreasing for this specific function?
4. What happens to $f(x)$ as $x \to \infty$?
5. How can we graphically interpret the interval where the function is increasing?

Tip: Always check the sign of the derivative to determine if a function is increasing or decreasing.