To identify the critical points of the function:

$y = \frac{3x}{x^2 - 25},$

we need to find where the derivative $y'$ is equal to zero or undefined. Critical points occur at values of $x$ where $y' = 0$ or $y'$ does not exist, as long as those $x$-values are within the domain of the original function.

Step 1: Find the Derivative

The function is a quotient, so we will use the quotient rule:

$y = \frac{u}{v}, \quad \text{where} \quad u = 3x \quad \text{and} \quad v = x^2 - 25.$

The quotient rule states:

$y' = \frac{u'v - uv'}{v^2},$

where: $u' = 3 \quad \text{and} \quad v' = 2x.$

Substitute these into the derivative formula:

$y' = \frac{(3)(x^2 - 25) - (3x)(2x)}{(x^2 - 25)^2}.$

Simplify the numerator:

$y' = \frac{3x^2 - 75 - 6x^2}{(x^2 - 25)^2} = \frac{-3x^2 - 75}{(x^2 - 25)^2}.$

$y' = \frac{-3(x^2 + 25)}{(x^2 - 25)^2}.$

Step 2: Identify Critical Points

The critical points are found by setting $y' = 0$ or where the derivative is undefined:

1. When $y' = 0$: The numerator must be zero:

   $-3(x^2 + 25) = 0.$

   Solving this, we find:

   $x^2 + 25 = 0.$

   Since $x^2 + 25 = 0$ has no real solutions, there are no critical points from setting $y' = 0$.

2. Where $y'$ is undefined: The derivative is undefined when the denominator is zero:

   $(x^2 - 25)^2 = 0 \Rightarrow x^2 - 25 = 0.$

   Solving this gives:

   $x^2 = 25 \Rightarrow x = \pm 5.$

   However, we must check the domain of the original function. The original function is undefined at $x = \pm 5$, so these are not critical points but rather vertical asymptotes.

Conclusion

There are no real critical points for this function. The points $x = \pm 5$ are where the function has vertical asymptotes, but there are no values of $x$ that make the derivative zero within the domain.

Would you like a detailed explanation of any step or have further questions? Here are some related questions to consider:

1. What are vertical asymptotes, and how are they different from critical points?
2. How can we find horizontal asymptotes for this function?
3. What does the quotient rule state, and why is it useful for derivatives?
4. What is the importance of checking the domain when finding critical points?
5. How would the process change if the function had a different denominator?

Tip: When dealing with rational functions, always check where the denominator is zero to determine where the function and its derivative might be undefined.