https://file.aiquickdraw.com/mathbot/file/1726158389401b8k7r3fa.jpg

To solve the problems on the image, we need to analyze the given function:

\[
f(x) = x^3(x + 5)^2
\]

### 1. Find the intervals where the function is increasing

To determine where the function is increasing or decreasing, we need to find its first derivative, \( f'(x) \), and analyze its sign.

\[
f(x) = x^3(x + 5)^2
\]

Applying the product rule:

\[
f'(x) = 3x^2(x + 5)^2 + x^3 \cdot 2(x + 5) \cdot 1
\]

Simplify the derivative:

\[
f'(x) = 3x^2(x + 5)^2 + 2x^3(x + 5)
\]
\[
f'(x) = x^2(x + 5)[3(x + 5) + 2x]
\]
\[
f'(x) = x^2(x + 5)(3x + 15 + 2x)
\]
\[
f'(x) = x^2(x + 5)(5x + 15)
\]

Setting \( f'(x) = 0 \), we find the critical points:

1. \( x^2 = 0 \)  ⟹ \( x = 0 \)
2. \( x + 5 = 0 \)  ⟹ \( x = -5 \)
3. \( 5x + 15 = 0 \)  ⟹ \( x = -3 \)

These critical points split the number line into four intervals: \( (-\infty, -5) \), \( (-5, -3) \), \( (-3, 0) \), and \( (0, \infty) \).

To determine where \( f'(x) \) is positive or negative, we test a value from each interval:

- For \( x = -6 \): \( f'(-6) = (-6)^2(-6 + 5)(5(-6) + 15) > 0 \) (Positive)
- For \( x = -4 \): \( f'(-4) = (-4)^2(-4 + 5)(5(-4) + 15) < 0 \) (Negative)
- For \( x = -1 \): \( f'(-1) = (-1)^2(-1 + 5)(5(-1) + 15) > 0 \) (Positive)
- For \( x = 1 \): \( f'(1) = (1)^2(1 + 5)(5(1) + 15) > 0 \) (Positive)

### Conclusion for increasing intervals:
- The function is increasing on the intervals \( (-\infty, -5) \), \( (-3, \infty) \).

### 2. Find the interval where the function is positive

We need to solve \( f(x) > 0 \). Since \( x^3(x + 5)^2 > 0 \), and \((x + 5)^2\) is always non-negative:

1. \( x^3 > 0 \) ⟹ \( x > 0 \)
2. The function is positive on the interval \( (0, 11] \).

### 3. Where does the function achieve its minimum?

To find the minimum, we examine the critical points and the endpoints of the interval \( x \in [-13, 11] \):

- At \( x = -5 \), \( f(-5) = (-5)^3 \cdot 0 = 0 \)
- The minimum occurs at \( x = -5 \).

### Summary of Answers:
1. **Increasing intervals:** \( (-\infty, -5) \), \( (-3, \infty) \).
2. **Interval where function is positive:** \( (0, 11] \).
3. **Function achieves minimum at:** \( x = -5 \).

If you have any questions or need further details, feel free to ask!

### Follow-Up Questions:
1. How do we find the second derivative to analyze concavity?
2. What are the critical points and why are they important?
3. How can we confirm the local minimum or maximum of the function?
4. What happens to the function as \( x \to \infty \) or \( x \to -\infty \)?
5. How do we interpret the intervals where the function is negative?

### Tip:
To fully understand a function's behavior, consider both the first and second derivatives for a complete analysis of its increasing/decreasing and concavity properties.

On which two intervals is the function increasing? Where does the function achieve its minimum? Find the interval on which the function is positive.

Learn how to find the intervals where a cubic function is increasing, determine where it achieves its minimum, and identify the intervals on which it is positive. This problem involves calculus concepts such as first derivatives and critical points. Suitable for high school students in Grades 10-12.

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