equation for function where f(-4)=f(1)=0; f’(-1.5)=0, f’(x)<0 for x < -1.5; f’(x) > 0 for x >-1.5

We are given the following information about the function \( f(x) \):

1. \( f(-4) = 0 \) and \( f(1) = 0 \), which means the points \( (-4, 0) \) and \( (1, 0) \) are roots of the function.
2. \( f'(-1.5) = 0 \), meaning \( f(x) \) has a critical point at \( x = -1.5 \).
3. \( f'(x) < 0 \) for \( x < -1.5 \), meaning the function is decreasing to the left of \( x = -1.5 \).
4. \( f'(x) > 0 \) for \( x > -1.5 \), meaning the function is increasing to the right of \( x = -1.5 \).

This suggests that \( f(x) \) has a **local minimum** at \( x = -1.5 \), since the derivative changes sign from negative to positive at this point. Also, \( f(x) \) is a polynomial with roots at \( x = -4 \) and \( x = 1 \), and a critical point at \( x = -1.5 \).

### General Form of the Function

Given the information, a cubic function is a good candidate because it can have one local extremum and two roots. So, we assume the function takes the form:
\[
f(x) = a(x + 4)(x - 1)(x + 1.5)
\]
where \( a \) is a constant to be determined.

To ensure \( f'(x) \) meets the conditions, we can expand and analyze this expression.

### Expansion:

Let's expand the polynomial first and analyze the derivative.

Would you like to proceed with this expansion, or do you want me to explore another type of function?

Explore how to determine a cubic function with roots at x = -4 and x = 1, and a critical point at x = -1.5. This problem covers polynomial behavior, derivative analysis, and critical points, suitable for students in Grades 10-12.

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