To determine where the function $f(x) = 2.638x^3 - 8.12x - 6.59$ changes from increasing to decreasing, we need to find the critical points by following these steps:

1. Find the first derivative $f'(x)$.
2. Set the derivative equal to zero to find the critical points.
3. Determine where the derivative changes sign to identify whether the function is increasing or decreasing.

Step 1: Find the first derivative

The first derivative of the function $f(x) = 2.638x^3 - 8.12x - 6.59$ is:

$f'(x) = 3 \cdot 2.638x^2 - 8.12$ $f'(x) = 7.914x^2 - 8.12$

Step 2: Set the derivative equal to zero

$7.914x^2 - 8.12 = 0$

Solve for $x$:

$7.914x^2 = 8.12$ $x^2 = \frac{8.12}{7.914}$ $x^2 = 1.026$ $x = \pm \sqrt{1.026}$ $x = \pm 1.013$

Thus, the critical points are $x = 1.013$ and $x = -1.013$.

Step 3: Determine where the derivative changes sign

To check where the function changes from increasing to decreasing, we need to evaluate the sign of $f'(x)$ around these critical points:

• For $x < -1.013$, $f'(x)$ will be positive (since $x^2$ term is positive).
• For $-1.013 < x < 1.013$, $f'(x)$ is negative.
• For $x > 1.013$, $f'(x)$ is positive again.

Thus, the function changes from increasing to decreasing at $x = -1.013$.

Would you like further details on this, or do you have any questions?

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Here are five related questions to explore:

1. What is the second derivative of $f(x)$, and what does it tell us?
2. How do you determine the intervals of concavity for this function?
3. Can you find the local maximum and minimum points for the function?
4. How does the cubic term affect the overall shape of the function?
5. What is the behavior of $f(x)$ as $x \to \infty$ and $x \to -\infty$?

Tip: To confirm if a point is a local maximum or minimum, use the second derivative test.