To solve this problem, we need to find a cubic polynomial $f(x) = ax^3 + bx^2 + cx + d$ that has horizontal tangents at the points $(-5, 5)$ and $(2, -5)$. Here are the steps to solve this problem:

Step 1: Understand Horizontal Tangents

A horizontal tangent at a point on the function $f(x)$ implies that the derivative of $f(x)$, $f'(x)$, is zero at that point.

Step 2: Find $f'(x)$

For the function $f(x) = ax^3 + bx^2 + cx + d$, the derivative $f'(x)$ is: $f'(x) = 3ax^2 + 2bx + c$

Step 3: Apply Horizontal Tangent Conditions

Since $f(x)$ has horizontal tangents at $x = -5$ and $x = 2$, we set $f'(-5) = 0$ and $f'(2) = 0$.

For $f'(-5) = 0$:

$3a(-5)^2 + 2b(-5) + c = 0$ Simplifying, we get: $75a - 10b + c = 0$

For $f'(2) = 0$:

$3a(2)^2 + 2b(2) + c = 0$ Simplifying, we get: $12a + 4b + c = 0$

Step 4: Apply Function Values

Since $f(-5) = 5$ and $f(2) = -5$, we have two more equations.

For $f(-5) = 5$:

$a(-5)^3 + b(-5)^2 + c(-5) + d = 5$ Simplifying, we get: $-125a + 25b - 5c + d = 5$

For $f(2) = -5$:

$a(2)^3 + b(2)^2 + c(2) + d = -5$ Simplifying, we get: $8a + 4b + 2c + d = -5$

Step 5: Solve the System of Equations

Now we have a system of four equations:

1. $75a - 10b + c = 0$
2. $12a + 4b + c = 0$
3. $-125a + 25b - 5c + d = 5$
4. $8a + 4b + 2c + d = -5$

I’ll solve this system to find $a$, $b$, $c$, and $d$.The solution to the system of equations is:

$a = \frac{20}{343}, \quad b = \frac{90}{343}, \quad c = -\frac{600}{343}, \quad d = -\frac{1035}{343}$

Thus, the cubic polynomial $f(x)$ that has horizontal tangents at the points $(-5, 5)$ and $(2, -5)$ is:

$f(x) = \frac{20}{343}x^3 + \frac{90}{343}x^2 - \frac{600}{343}x - \frac{1035}{343}$

Would you like further details on the solution process, or do you have any questions?

Here are five questions related to this topic that might help deepen your understanding:

1. How can you verify if a cubic polynomial has horizontal tangents at specific points?
2. Why does setting $f'(x) = 0$ at certain points ensure horizontal tangents?
3. What role do the specific points $(-5,5)$ and $(2,-5)$ play in determining the constants $a$, $b$, $c$, and $d$?
4. How would the solution change if the given points had different values?
5. Can you generalize this approach to find a cubic polynomial with horizontal tangents at any two given points?

Tip: Solving systems of equations is a common approach to finding unknown coefficients in polynomial equations, especially when derivatives and function values at specific points are involved.