Let's solve each part of the problem step-by-step.

1. Determine $A$ where the function is undefined

The function given is: $f(x) = \frac{3x + 9}{4x + 3}$

The function is undefined where the denominator equals zero. Thus, we solve: $4x + 3 = 0$ $4x = -3$ $x = -\frac{3}{4}$

So, $A = -\frac{3}{4}$.

2. Determine whether $f(x)$ is increasing or decreasing on each interval

To determine whether $f(x)$ is increasing or decreasing on the intervals $(-\infty, A)$ and $(A, \infty)$, we find the derivative $f'(x)$ and examine its sign.

Finding $f'(x)$

Using the quotient rule for derivatives, where $f(x) = \frac{u}{v}$ with $u = 3x + 9$ and $v = 4x + 3$: $f'(x) = \frac{v \cdot u' - u \cdot v'}{v^2}$

• $u' = 3$
• $v' = 4$

Thus, $f'(x) = \frac{(4x + 3)(3) - (3x + 9)(4)}{(4x + 3)^2}$ Simplifying, $f'(x) = \frac{12x + 9 - 12x - 36}{(4x + 3)^2}$ $f'(x) = \frac{-27}{(4x + 3)^2}$

Since $(4x + 3)^2$ is always positive, $f'(x)$ will always be negative (because the numerator $-27$ is negative). This implies that $f(x)$ is always decreasing on both intervals.

• On $(-\infty, A)$: Decreasing
• On $(A, \infty)$: Decreasing

3. Determine the concavity of $f(x)$ on each interval

To determine concavity, we need the second derivative $f''(x)$ and its sign.

Finding $f''(x)$

To differentiate $f'(x) = \frac{-27}{(4x + 3)^2}$ again, we can use the chain rule: $f''(x) = \frac{d}{dx} \left( -27 \cdot (4x + 3)^{-2} \right)$ Letting $g(x) = (4x + 3)^{-2}$, we find $g'(x)$: $g'(x) = -2(4x + 3)^{-3} \cdot 4 = -8(4x + 3)^{-3}$ Thus, $f''(x) = -27 \cdot \left(-8(4x + 3)^{-3}\right) = 216(4x + 3)^{-3}$

Since $(4x + 3)^{-3}$ is positive for $x > -\frac{3}{4}$ and negative for $x < -\frac{3}{4}$, we find:

• On $(-\infty, A)$: Concave Down
• On $(A, \infty)$: Concave Up

Summary

• Value of $A$: $A = -\frac{3}{4}$
• Monotonicity:
  • $(-\infty, A)$: Decreasing
  • $(A, \infty)$: Decreasing
• Concavity:
  • $(-\infty, A)$: Concave Down
  • $(A, \infty)$: Concave Up

Would you like a deeper explanation on any step, or do you have questions?

Related Questions:

1. What is the significance of a function's concavity in understanding its graph?
2. How can we interpret the behavior of a function around a vertical asymptote?
3. Why is it helpful to know where a function is increasing or decreasing?
4. What is the importance of finding intervals of concavity for real-world applications?
5. How does the derivative inform us about the function's slope and behavior?

Tip:

When analyzing rational functions, always start by identifying points of discontinuity (where the denominator is zero) to understand the function's domain.