To find the concavity and inflection points of the function $f(x) = \frac{x^2 - 3x + 9}{x - 3}$, we follow these steps:

1. Domain of the function:

The function is undefined at $x = 3$ because of the denominator $(x - 3)$. So, the domain of the function is $x \in (-\infty, 3) \cup (3, \infty)$.

2. First derivative ( $f'(x)$ ):

We need the first derivative to find the critical points and intervals where the function is increasing or decreasing. Using the quotient rule for derivatives:

$f'(x) = \frac{(x - 3) \cdot (2x - 3) - (x^2 - 3x + 9) \cdot 1}{(x - 3)^2}$

Simplify the numerator:

$f'(x) = \frac{(2x^2 - 6x) - (x^2 - 3x + 9)}{(x - 3)^2}$ $f'(x) = \frac{2x^2 - 6x - x^2 + 3x - 9}{(x - 3)^2}$ $f'(x) = \frac{x^2 - 3x - 9}{(x - 3)^2}$

3. Second derivative ( $f''(x)$ ):

To determine concavity, we need the second derivative. Again, use the quotient rule on $f'(x)$:

$f''(x) = \frac{(x - 3)^2 \cdot (2x - 3) - (x^2 - 3x - 9) \cdot 2(x - 3)}{(x - 3)^4}$

Simplifying the numerator:

$f''(x) = \frac{(x - 3)^2 \cdot (2x - 3) - 2(x^2 - 3x - 9)(x - 3)}{(x - 3)^4}$

4. Concavity and Inflection Points:

To determine concavity:

• Solve $f''(x) = 0$ to find potential inflection points.
• Analyze the sign of $f''(x)$ to determine where the function is concave up ( $f''(x) > 0$ ) or concave down ( $f''(x) < 0$ ).

Let me calculate these details for you!The first and second derivatives of the function $f(x) = \frac{x^2 - 3x + 9}{x - 3}$ are:

1. First derivative: $f'(x) = \frac{x(x - 6)}{(x - 3)^2}$

2. Second derivative: $f''(x) = \frac{18}{(x - 3)^3}$

Concavity and Inflection Points:

• Inflection Points:
  The inflection points occur where the second derivative $f''(x) = 0$. However, since the second derivative $\frac{18}{(x - 3)^3}$ is never zero, there are no real inflection points.

• Concavity:

  • Concave Up: When $f''(x) > 0$, the function is concave up. Since $f''(x) = \frac{18}{(x - 3)^3}$, it is positive when $x > 3$.
  • Concave Down: When $f''(x) < 0$, the function is concave down. It is negative when $x < 3$.

Summary:

• The function is concave down on the interval $(-\infty, 3)$.
• The function is concave up on the interval $(3, \infty)$.
• There are no inflection points, as the second derivative never equals zero.

Would you like further details on this process or more examples?

Here are 5 related questions:

1. What are the critical points of the function?
2. How would you find the vertical asymptotes of the function?
3. How can we verify the concavity using a graphing tool?
4. What happens to the behavior of the function near $x = 3$?
5. How do you analyze concavity for more complex functions?

Tip: Always check for points where the denominator of the derivative becomes zero, as they often indicate important features such as asymptotes or undefined regions.