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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.

Find a cubic polynomial f(x) = ax^3 + bx^2 + cx + d that has horizontal tangents at the points (-5, 5) and (2, -5).

This page provides a detailed solution 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). Learn how to use derivatives and solve systems of equations to determine the coefficients.

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