To find the critical numbers of the function $f(x) = \frac{4x}{x^2 + 1}$, we first need to find its derivative and determine where it is equal to zero or undefined. These points are the critical numbers.

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

We'll use the quotient rule to differentiate. The quotient rule states:

$\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot u' - u \cdot v'}{v^2}$

For $f(x) = \frac{4x}{x^2 + 1}$, let:

• $u = 4x$, so $u' = 4$,
• $v = x^2 + 1$, so $v' = 2x$.

Now, apply the quotient rule:

$f'(x) = \frac{(x^2 + 1)(4) - (4x)(2x)}{(x^2 + 1)^2}$

Simplify the numerator:

$f'(x) = \frac{4(x^2 + 1) - 8x^2}{(x^2 + 1)^2}$

$f'(x) = \frac{4x^2 + 4 - 8x^2}{(x^2 + 1)^2}$

$f'(x) = \frac{-4x^2 + 4}{(x^2 + 1)^2}$

$f'(x) = \frac{4(1 - x^2)}{(x^2 + 1)^2}$

Step 2: Set the derivative equal to zero

To find the critical numbers, we need to solve $f'(x) = 0$:

$\frac{4(1 - x^2)}{(x^2 + 1)^2} = 0$

The fraction equals zero when the numerator is zero, so set the numerator $4(1 - x^2) = 0$:

$1 - x^2 = 0$

$x^2 = 1$

$x = \pm 1$

Step 3: Check for undefined points

Since the denominator $(x^2 + 1)^2$ is never zero for any real $x$, the derivative is never undefined.

Conclusion

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

Would you like more details on finding critical numbers or help with graphing this function?

Here are 5 related questions:

1. How do we classify critical points as maxima, minima, or points of inflection?
2. What is the behavior of the function $f(x) = \frac{4x}{x^2 + 1}$ near $x = 1$ and $x = -1$?
3. How do we analyze the concavity of a function using the second derivative?
4. How do we find horizontal asymptotes of rational functions?
5. How do we compute the global maximum or minimum of $f(x) = \frac{4x}{x^2 + 1}$?

Tip: When solving optimization problems, always verify critical points using the first or second derivative test to confirm if they are maxima or minima.