Sketch a graph of g(x). g’(x) = ((4-x)/(6-x)^⅔ * x^⅓)) g’’(x) = ((-8)/(6-x)^5/3 * x^4/3)
 


To sketch the graph of the function \( g(x) \), we can analyze its behavior using the given first and second derivatives:

1. **First derivative \( g'(x) \):**

\[
g'(x) = \frac{4-x}{(6-x)^{\frac{2}{3}} \cdot x^{\frac{1}{3}}}
\]
The first derivative tells us about the slope of the function and where it is increasing or decreasing.

- **Critical points** are found where \( g'(x) = 0 \) or \( g'(x) \) is undefined. Solving \( g'(x) = 0 \), we set the numerator equal to zero:

\[
4 - x = 0 \quad \Rightarrow \quad x = 4
\]

- The denominator is undefined when \( x = 0 \) or \( x = 6 \), so we need to be cautious about these points in our analysis.

2. **Second derivative \( g''(x) \):**

\[
g''(x) = \frac{-8}{(6-x)^{\frac{5}{3}} \cdot x^{\frac{4}{3}}}
\]
The second derivative helps us analyze the concavity of the function and locate inflection points.

- The second derivative is undefined when \( x = 0 \) and \( x = 6 \), just like the first derivative. Since \( g''(x) \) does not change sign at these points, they may correspond to vertical asymptotes.

3. **Behavior at specific points:**

- **At \( x = 0 \)**: Both \( g'(x) \) and \( g''(x) \) are undefined, indicating a potential vertical asymptote.
- **At \( x = 4 \)**: Since \( g'(4) = 0 \), we have a critical point at \( x = 4 \). To determine if it is a maximum or minimum, we check \( g''(x) \):

\[
g''(4) = \frac{-8}{(6-4)^{\frac{5}{3}} \cdot 4^{\frac{4}{3}}} < 0
\]
Since \( g''(4) < 0 \), this means \( x = 4 \) is a local maximum.

- **At \( x = 6 \)**: The function has another vertical asymptote because both \( g'(x) \) and \( g''(x) \) are undefined.

### Sketch of \( g(x) \):

- **Asymptotes**: Vertical asymptotes at \( x = 0 \) and \( x = 6 \).
- **Critical points**: A local maximum at \( x = 4 \).
- **Concavity**: The function is concave down near \( x = 4 \) due to \( g''(x) < 0 \).

### Want an image graph sketch? If so, I'll generate one!

Let me know if you need any details or have questions! Here are some related questions to consider:

1. How does the behavior of \( g'(x) \) near \( x = 0 \) and \( x = 6 \) affect the overall graph shape?
2. What happens to the graph as \( x \to 0 \) from the right?
3. How do we interpret the concavity changes based on \( g''(x) \)?
4. Can we infer more about the end behavior of \( g(x) \)?
5. Are there any inflection points based on the second derivative?

**Tip**: Remember, vertical asymptotes often indicate where the function heads toward infinity or negative infinity.

Sketch a graph of g(x). g’(x) = ((4-x)/(6-x)^⅔ * x^⅓)) g’’(x) = ((-8)/(6-x)^5/3 * x^4/3)

Analyze the function g(x) using its first and second derivatives to sketch its graph. Explore critical points, asymptotes, and concavity changes with a detailed breakdown. Learn how g'(x) and g''(x) influence the graph of g(x) at x = 4 and around vertical asymptotes.

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