To find the range of values of $x$ for which the given curve $y = \frac{x+1}{x^2 + 3}$ is a decreasing function, we need to follow these steps:

Steps:

1. Differentiate $y$: Compute $\frac{dy}{dx}$, the derivative of $y$, using the quotient rule: $\frac{dy}{dx} = \frac{(x^2 + 3) \cdot \frac{d}{dx}(x+1) - (x+1) \cdot \frac{d}{dx}(x^2 + 3)}{(x^2 + 3)^2}.$

2. Simplify $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{(x^2 + 3)(1) - (x+1)(2x)}{(x^2 + 3)^2}.$ Expanding: $\frac{dy}{dx} = \frac{x^2 + 3 - 2x^2 - 2x}{(x^2 + 3)^2}.$ Combine like terms: $\frac{dy}{dx} = \frac{-x^2 - 2x + 3}{(x^2 + 3)^2}.$

3. Find when $\frac{dy}{dx} < 0$: A function is decreasing when $\frac{dy}{dx} < 0$. So, solve: $-x^2 - 2x + 3 < 0.$ Factorize $-x^2 - 2x + 3$: $-(x^2 + 2x - 3) < 0.$ Simplify further: $x^2 + 2x - 3 > 0.$ Factorize $x^2 + 2x - 3$: $(x + 3)(x - 1) > 0.$

4. Solve the inequality $(x + 3)(x - 1) > 0$: The critical points are $x = -3$ and $x = 1$. Use a number line test:

   • For $x < -3$, both factors $(x + 3)$ and $(x - 1)$ are negative, so the product is positive.
   • For $-3 < x < 1$, $(x + 3) > 0$ and $(x - 1) < 0$, so the product is negative.
   • For $x > 1$, both factors $(x + 3)$ and $(x - 1)$ are positive, so the product is positive.

   Thus, $(x + 3)(x - 1) > 0$ for $x < -3$ or $x > 1$.

5. Conclusion: The curve $y = \frac{x+1}{x^2 + 3}$ is decreasing for: $-3 < x < 1.$

Would you like more details or further explanation on any part of this?

Here are 5 related questions:

1. How do you interpret the critical points $x = -3$ and $x = 1$ in this context?
2. What does the sign of the derivative tell us about increasing and decreasing intervals?
3. How would the analysis change if the denominator of the function $y$ was different?
4. Can this function have stationary points, and how can they be determined?
5. How can the behavior of this function be visualized on a graph?

Tip: Always analyze the sign of the derivative over intervals, not just at the critical points!