The uploaded image appears to be a mathematical problem in handwritten form. To solve it, I will extract and interpret the content step by step.

The main function provided in the text is:

$f(x) = \frac{3(x-1)^3}{3x^3 + 1}.$

Key Tasks Extracted from the Image:

1. Domain of the Function $f(x)$:

   • Identify values of $x$ where the denominator becomes zero since the function is undefined at those points.

2. Intercepts of the Function:

   • Find the $y$-intercept by evaluating $f(0)$.
   • Find the $x$-intercept(s) by solving $f(x) = 0$, which occurs when the numerator is zero.

3. Behavior as $x \to \pm\infty$:

   • Analyze the asymptotic behavior of $f(x)$ as $x$ approaches positive and negative infinity.

4. Critical Points:

   • Calculate $f'(x)$ and find where it is zero or undefined to determine critical points.
   • Use these to identify intervals of increase and decrease.

5. Inflection Points:

   • Compute $f''(x)$ to find points of inflection and intervals of concavity.

6. Graphing the Function:

   • Sketch the function $f(x)$ based on the above information.

Let's Solve Each Step:

1. Domain of $f(x)$:

The denominator is $3x^3 + 1$. Setting this equal to zero: $3x^3 + 1 = 0 \quad \implies \quad x^3 = -\frac{1}{3} \quad \implies \quad x = -\sqrt[3]{\frac{1}{3}}.$ Thus, $f(x)$ is undefined at $x = -\sqrt[3]{\frac{1}{3}}$.

2. Intercepts:

• $y$-Intercept: Substitute $x = 0$ into $f(x)$: $f(0) = \frac{3(0-1)^3}{3(0)^3 + 1} = \frac{3(-1)^3}{1} = -3.$ Thus, the $y$-intercept is $(0, -3)$.

• $x$-Intercept(s): The numerator is $3(x-1)^3$. Setting this equal to zero: $3(x-1)^3 = 0 \quad \implies \quad x-1 = 0 \quad \implies \quad x = 1.$ Thus, the $x$-intercept is $(1, 0)$.

3. Behavior as $x \to \pm\infty$:

For large values of $x$, the highest degree terms dominate: $f(x) \approx \frac{3x^3}{3x^3} = 1.$ Thus, the horizontal asymptote is $y = 1$.

4. Critical Points:

To find critical points, calculate $f'(x)$ using the quotient rule: $f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2},$ where $g(x) = 3(x-1)^3$ and $h(x) = 3x^3 + 1$.

After calculating: $f'(x) = \frac{9(x-1)^2(3x^3 + 1) - 9x^2 \cdot 3(x-1)^3}{(3x^3 + 1)^2}.$ Simplify and solve $f'(x) = 0$ to find critical points.

5. Inflection Points:

Compute $f''(x)$ and set it to zero. This involves applying the product and quotient rules again.

6. Sketch the Graph:

Use all the above results, including intercepts, asymptotes, and concavity, to plot the graph of $f(x)$.

Would you like me to provide a detailed calculation for steps 4 and 5, or sketch the graph? Let me know!

Related Questions:

1. What is the importance of identifying the domain of a function?
2. How does asymptotic behavior help in sketching graphs of rational functions?
3. Why do we use the derivative to determine critical points and intervals of increase/decrease?
4. What is the significance of finding inflection points in a graph?
5. How do horizontal asymptotes differ from vertical asymptotes?

Tip:

When solving for critical or inflection points, always double-check your differentiation process to avoid errors!