To find the critical numbers of the given function, $f(x) = \frac{3x+3}{5x^2 + 5x + 5}$, we follow these steps:

Step 1: Find the derivative of the function.

The derivative is found using the quotient rule: $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}$ Here, $u = 3x + 3$ and $v = 5x^2 + 5x + 5$.

• $u' = 3$
• $v' = 10x + 5$

Substituting into the quotient rule: $f'(x) = \frac{(3)(5x^2 + 5x + 5) - (3x + 3)(10x + 5)}{(5x^2 + 5x + 5)^2}$

Simplify the numerator: $f'(x) = \frac{15x^2 + 15x + 15 - [(30x^2 + 15x) + (30x + 15)]}{(5x^2 + 5x + 5)^2}$ $f'(x) = \frac{15x^2 + 15x + 15 - 30x^2 - 45x - 15}{(5x^2 + 5x + 5)^2}$ $f'(x) = \frac{-15x^2 - 30x}{(5x^2 + 5x + 5)^2}$

Step 2: Set the derivative equal to zero.

For critical numbers, $f'(x) = 0$. The numerator must be zero: $-15x^2 - 30x = 0$ Factorize: $-15x(x + 2) = 0$ $x = 0 \quad \text{or} \quad x = -2$

Step 3: Exclude values that make the denominator zero.

The denominator $(5x^2 + 5x + 5)^2$ is never zero because $5x^2 + 5x + 5 = 0$ has no real roots (the discriminant is negative). Thus, there are no restrictions here.

Critical Numbers:

The critical numbers are: $x = 0 \quad \text{and} \quad x = -2$

Let me know if you would like a detailed explanation or further verification of the steps!

Here are 5 questions to deepen your understanding:

1. How does the quotient rule apply to derivatives of rational functions?
2. Why is it necessary to check if the denominator of the function is zero?
3. How do critical numbers relate to extrema or inflection points?
4. Can a critical number exist outside the domain of the function?
5. What is the significance of the discriminant in analyzing the denominator?

Tip: Always double-check for potential restrictions on the domain when finding critical numbers of rational functions.