To find the critical numbers of the function $f(x) = \frac{4x}{x^2 + 1}$, we need to:

1. Find the derivative of $f(x)$.
2. Set the derivative equal to zero and solve for $x$.
3. Identify where the derivative does not exist, if applicable, since those points may also yield critical numbers.

Step 1: Differentiate $f(x)$

Using the quotient rule, where if $f(x) = \frac{g(x)}{h(x)}$, then $f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}$, we can differentiate $f(x) = \frac{4x}{x^2 + 1}$.

1. Let $g(x) = 4x$, so $g'(x) = 4$.
2. Let $h(x) = x^2 + 1$, so $h'(x) = 2x$.

Using the quotient rule: $f'(x) = \frac{(4)(x^2 + 1) - (4x)(2x)}{(x^2 + 1)^2}$ Simplify the numerator: $f'(x) = \frac{4x^2 + 4 - 8x^2}{(x^2 + 1)^2} = \frac{-4x^2 + 4}{(x^2 + 1)^2}$ Factor out a $-4$ from the numerator: $f'(x) = \frac{-4(x^2 - 1)}{(x^2 + 1)^2} = \frac{-4(x + 1)(x - 1)}{(x^2 + 1)^2}$

Step 2: Set the derivative equal to zero

To find the critical points, set $f'(x) = 0$: $\frac{-4(x + 1)(x - 1)}{(x^2 + 1)^2} = 0$ The fraction is zero when the numerator is zero, so: $-4(x + 1)(x - 1) = 0$ This gives $x = 1$ and $x = -1$ as solutions.

Step 3: Check where $f'(x)$ is undefined

The derivative $f'(x)$ is undefined when $(x^2 + 1)^2 = 0$. However, $x^2 + 1 = 0$ has no real solutions, so $f'(x)$ is defined for all real $x$.

Conclusion

The critical numbers of $f(x) = \frac{4x}{x^2 + 1}$ are: $x = 1 \quad \text{and} \quad x = -1$

Would you like further details or have any questions?

Here are 5 additional questions to consider:

1. How would you find the relative extrema at the critical points?
2. What is the behavior of $f(x)$ as $x \to \infty$?
3. Can we analyze concavity to determine points of inflection for $f(x)$?
4. How does the behavior of $f(x)$ compare near $x = 0$?
5. What role does the denominator $x^2 + 1$ play in determining the domain of $f(x)$?

Tip: When working with derivatives of fractions, the quotient rule is essential for simplifying expressions accurately.