The question asks to analyze the function $f(x) = x^3(1-x)$ through several mathematical aspects. Here's a step-by-step solution:

1. Determine the degree of the function:

The function $f(x) = x^3(1-x)$ simplifies as follows: $f(x) = x^3 - x^4$ This is a polynomial of degree 4 because the highest power of $x$ is 4.

• Answer: The degree of the function is 4.

2. Find the intervals of increase and decrease:

To find intervals of increase and decrease, we compute the derivative $f'(x)$: $f'(x) = \frac{d}{dx}(x^3 - x^4) = 3x^2 - 4x^3$ Factorizing: $f'(x) = x^2(3 - 4x)$ Setting $f'(x) = 0$: $x^2(3 - 4x) = 0$ Solutions: $x = 0 \quad \text{or} \quad x = \frac{3}{4}$ These critical points divide the domain into intervals: $(-\infty, 0)$, $(0, \frac{3}{4})$, and $(\frac{3}{4}, \infty)$.

Test the sign of $f'(x)$ in each interval:

• For $x \in (-\infty, 0)$, $f'(x) > 0$ (increasing).

• For $x \in (0, \frac{3}{4})$, $f'(x) > 0$ (increasing).

• For $x \in (\frac{3}{4}, \infty)$, $f'(x) < 0$ (decreasing).

• Answer:

  • Increasing: $(-\infty, \frac{3}{4})$
  • Decreasing: $(\frac{3}{4}, \infty)$

3. Find the maximum and minimum values:

Using the critical points $x = 0$ and $x = \frac{3}{4}$, calculate $f(x)$:

• At $x = 0$: $f(0) = 0^3(1-0) = 0$
• At $x = \frac{3}{4}$: $f\left(\frac{3}{4}\right) = \left(\frac{3}{4}\right)^3 \left(1 - \frac{3}{4}\right) = \frac{27}{64} \cdot \frac{1}{4} = \frac{27}{256}$

Since the function decreases after $x = \frac{3}{4}$, the maximum value is $\frac{27}{256}$ at $x = \frac{3}{4}$, and the minimum is 0 at $x = 0$.

• Answer:
  • Maximum: $\frac{27}{256}$ at $x = \frac{3}{4}$
  • Minimum: 0 at $x = 0$

4. Find the inflection points:

The second derivative helps locate inflection points: $f''(x) = \frac{d}{dx}(3x^2 - 4x^3) = 6x - 12x^2$ Factorizing: $f''(x) = 6x(1 - 2x)$ Setting $f''(x) = 0$: $6x(1 - 2x) = 0$ Solutions: $x = 0 \quad \text{or} \quad x = \frac{1}{2}$ Test the concavity:

• For $x \in (-\infty, 0)$, $f''(x) < 0$ (concave down).

• For $x \in (0, \frac{1}{2})$, $f''(x) > 0$ (concave up).

• For $x \in (\frac{1}{2}, \infty)$, $f''(x) < 0$ (concave down).

• Answer: Inflection points at $x = 0$ and $x = \frac{1}{2}$.

5. Find regions of concavity:

Using the results from the second derivative:

• Concave up: $(0, \frac{1}{2})$

• Concave down: $(-\infty, 0) \cup (\frac{1}{2}, \infty)$

• Answer:

  • Concave up: $(0, \frac{1}{2})$
  • Concave down: $(-\infty, 0) \cup (\frac{1}{2}, \infty)$

Would you like a detailed plot or further clarification?

Related Questions:

1. How can the critical points be used to analyze the function further?
2. What is the role of inflection points in determining the behavior of $f(x)$?
3. How does the degree of the polynomial influence its general shape?
4. Can we approximate the maximum value using numerical methods instead of exact solutions?
5. How does increasing and decreasing behavior relate to real-world applications?

Tip:

Always verify the derivative computations when analyzing critical points or concavity, as small errors can propagate.