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The problem in the image involves finding the **point of inflection** and the **concavity** of the function \( f(x) = x \sqrt{6 - x} \).

### Step 1: Find the Point of Inflection
To find the point of inflection, we need to compute the second derivative of the function \( f(x) \) and determine where it changes sign (which could indicate an inflection point).

#### a) First Derivative \( f'(x) \)
We use the product rule and chain rule for differentiation:
\[
f(x) = x (6 - x)^{1/2}
\]
Using the product rule \( (uv)' = u'v + uv' \):
\[
f'(x) = \frac{d}{dx} \left( x \right) \cdot \sqrt{6 - x} + x \cdot \frac{d}{dx} \left( \sqrt{6 - x} \right)
\]
The derivative of \( \sqrt{6 - x} \) is:
\[
\frac{d}{dx} \left( \sqrt{6 - x} \right) = \frac{-1}{2 \sqrt{6 - x}}
\]
Thus:
\[
f'(x) = \sqrt{6 - x} + x \cdot \left( \frac{-1}{2 \sqrt{6 - x}} \right)
\]
Simplifying:
\[
f'(x) = \sqrt{6 - x} - \frac{x}{2 \sqrt{6 - x}}
\]
We can combine these terms by getting a common denominator:
\[
f'(x) = \frac{2(6 - x) - x}{2 \sqrt{6 - x}} = \frac{12 - 3x}{2 \sqrt{6 - x}}
\]

#### b) Second Derivative \( f''(x) \)
We now differentiate \( f'(x) = \frac{12 - 3x}{2 \sqrt{6 - x}} \). Using the quotient rule:
\[
f''(x) = \frac{(2 \sqrt{6 - x}) \cdot (-3) - (12 - 3x) \cdot \frac{-1}{2 \sqrt{6 - x}}}{(2 \sqrt{6 - x})^2}
\]
Simplifying the numerator:
\[
f''(x) = \frac{-6 \sqrt{6 - x} + (12 - 3x) \cdot \frac{1}{\sqrt{6 - x}}}{4(6 - x)}
\]
Simplifying further:
\[
f''(x) = \frac{-6(6 - x) + (12 - 3x)}{4(6 - x)^{3/2}}
\]
\[
f''(x) = \frac{-36 + 6x + 12 - 3x}{4(6 - x)^{3/2}} = \frac{-24 + 3x}{4(6 - x)^{3/2}}
\]

#### c) Set \( f''(x) = 0 \) to find inflection points
\[
\frac{-24 + 3x}{4(6 - x)^{3/2}} = 0
\]
This simplifies to:
\[
-24 + 3x = 0
\]
\[
3x = 24 \quad \Rightarrow \quad x = 8
\]
Since \( x = 8 \) is outside the domain of the function (because \( 6 - x \geq 0 \), which implies \( x \leq 6 \)), there is **no inflection point**. Therefore, the point of inflection **does not exist (DNE)**.

### Step 2: Describe the Concavity
To describe the concavity, we check the sign of \( f''(x) = \frac{-24 + 3x}{4(6 - x)^{3/2}} \).

- For concavity upwards, \( f''(x) > 0 \).
- For concavity downwards, \( f''(x) < 0 \).

#### a) Test the intervals:
- For \( x \in (0, 6) \), test \( x = 2 \):
  \[
  f''(2) = \frac{-24 + 3(2)}{4(6 - 2)^{3/2}} = \frac{-24 + 6}{4(4)^{3/2}} = \frac{-18}{32} < 0
  \]
  So, the function is concave **downward** on \( (0, 6) \).

Thus, the function is concave downward on \( (0, 6) \) and **DNE** for concave upward.

### Final Answer:
- **Point of inflection**: DNE
- **Concavity**:
  - Concave upward: DNE
  - Concave downward: \( (0, 6) \)

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Would you like any more details on these steps or clarifications?

Here are some related questions for further practice:
1. How do you compute the inflection points of a rational function?
2. Can a function have multiple points of inflection? Why or why not?
3. How does concavity change affect the shape of a function's graph?
4. What happens if the second derivative is undefined at some points?
5. What is the geometric interpretation of concavity on a graph?

**Tip**: Always ensure that the domain of a function is taken into account when solving for inflection points and concavity.

Find the point of inflection of the graph of the function f(x) = x sqrt(6 - x) and describe the concavity.

Learn how to find the point of inflection and describe the concavity of the function f(x) = x sqrt(6 - x). This problem explores advanced differentiation techniques, including the quotient rule, to find the second derivative and determine where the function changes concavity. Suitable for students in Grades 11-12 studying calculus.

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