given the function f(x)=2.413x^3+3.167x^2-8.559x+1.842 over the interval -4≤x≤3,
a. determine the interval(s) that f(x) is decreasing.

To determine where the function \( f(x) = 2.413x^3 + 3.167x^2 - 8.559x + 1.842 \) is decreasing over the interval \( -4 \leq x \leq 3 \), we need to follow these steps:

### Step 1: Find the derivative of \( f(x) \)
The derivative, \( f'(x) \), tells us the slope of the function at any point. When \( f'(x) < 0 \), the function is decreasing.

Let's compute \( f'(x) \):
\[
f'(x) = \frac{d}{dx}\left( 2.413x^3 + 3.167x^2 - 8.559x + 1.842 \right)
\]
Applying the power rule to each term:

\[
f'(x) = 3(2.413)x^2 + 2(3.167)x - 8.559
\]
\[
f'(x) = 7.239x^2 + 6.334x - 8.559
\]

### Step 2: Set \( f'(x) = 0 \) to find critical points
Next, solve the equation \( 7.239x^2 + 6.334x - 8.559 = 0 \) to find the critical points. These points indicate where the slope is zero, which helps identify potential changes in the behavior of the function (from increasing to decreasing or vice versa).

We'll solve this quadratic equation:

\[
7.239x^2 + 6.334x - 8.559 = 0
\]

Now I'll compute the roots of this quadratic.The critical points are \( x = 0.735 \) and \( x = -1.610 \). These are the points where the derivative is zero, which may indicate a change in the behavior of the function.

### Step 3: Determine the sign of \( f'(x) \) in each interval
The critical points divide the interval \( -4 \leq x \leq 3 \) into three sub-intervals:
1. \( -4 \leq x < -1.610 \)
2. \( -1.610 < x < 0.735 \)
3. \( 0.735 < x \leq 3 \)

To determine where the function is decreasing, we need to check the sign of \( f'(x) \) in each of these intervals. Let's evaluate \( f'(x) \) at test points within these intervals.

- Test point in \( (-4, -1.610) \): \( x = -3 \)
- Test point in \( (-1.610, 0.735) \): \( x = 0 \)
- Test point in \( (0.735, 3) \): \( x = 2 \)

I'll compute the sign of \( f'(x) \) at these points.The values of \( f'(x) \) at the test points are:

- \( f'(-3) = 37.59 \) (positive)
- \( f'(0) = -8.559 \) (negative)
- \( f'(2) = 33.065 \) (positive)

### Step 4: Conclusion
- In the interval \( (-4, -1.610) \), \( f'(x) > 0 \), so the function is **increasing**.
- In the interval \( (-1.610, 0.735) \), \( f'(x) < 0 \), so the function is **decreasing**.
- In the interval \( (0.735, 3) \), \( f'(x) > 0 \), so the function is **increasing**.

### Final Answer:
The function \( f(x) \) is decreasing on the interval \( (-1.610, 0.735) \).

---

Do you want further details, or do you have any questions? Here are some related questions to consider:

1. How do we classify the critical points of a function?
2. What is the significance of the sign of the second derivative?
3. How do we find inflection points in a cubic function?
4. How can you approximate the behavior of \( f(x) \) near critical points?
5. How can you use the first derivative test to confirm increasing/decreasing behavior?

**Tip:** The first derivative test helps you confirm whether a critical point is a local maximum, minimum, or neither based on the change in sign of \( f'(x) \).

Given the function f(x) = 2.413x^3 + 3.167x^2 - 8.559x + 1.842 over the interval -4 ≤ x ≤ 3, determine the interval(s) that f(x) is decreasing.

Determine where the function f(x) = 2.413x^3 + 3.167x^2 - 8.559x + 1.842 is decreasing over the interval -4 ≤ x ≤ 3. This solution involves using derivatives, solving a quadratic equation, and applying the first derivative test. Ideal for students at Grade 12 or College Level.

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