A particle moves along the *x*-axis so that at time t, is greater than or equal to, 0*t*≥0 its position is given by x, left parenthesis, t, right parenthesis, equals, t, cubed, minus, 6, t, squared, minus, 96, t, .*x*(*t*)=*t*3−6*t*2−96*t*. Determine all intervals when the speed of the particle is increasing.

To determine when the speed of the particle is increasing, we need to analyze the velocity and acceleration of the particle. The speed of the particle increases when the velocity and acceleration have the same sign (both positive or both negative).

### Step 1: Find the Velocity and Acceleration Functions

The position of the particle is given by:

\[
x(t) = t^3 - 6t^2 - 96t.
\]

#### Velocity Function
The velocity, \( v(t) \), is the derivative of the position function:

\[
v(t) = \frac{d}{dt} (t^3 - 6t^2 - 96t) = 3t^2 - 12t - 96.
\]

#### Acceleration Function
The acceleration, \( a(t) \), is the derivative of the velocity function:

\[
a(t) = \frac{d}{dt} (3t^2 - 12t - 96) = 6t - 12.
\]

### Step 2: Find Critical Points of Velocity and Acceleration

#### Critical Points of Velocity
To find when the velocity is zero, set \( v(t) = 0 \):

\[
3t^2 - 12t - 96 = 0.
\]

Divide the equation by 3:

\[
t^2 - 4t - 32 = 0.
\]

Solve using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\[
t = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-32)}}{2(1)} = \frac{4 \pm \sqrt{16 + 128}}{2} = \frac{4 \pm \sqrt{144}}{2} = \frac{4 \pm 12}{2}.
\]

This gives:

\[
t = 8 \quad \text{and} \quad t = -4.
\]

Since \( t \geq 0 \), we only consider \( t = 8 \).

#### Critical Points of Acceleration
Set \( a(t) = 0 \):

\[
6t - 12 = 0 \implies t = 2.
\]

### Step 3: Determine the Signs of Velocity and Acceleration

To determine when the speed is increasing, check the signs of \( v(t) \) and \( a(t) \) around the critical points \( t = 2 \) and \( t = 8 \).

1. **Interval \( (0, 2) \):**
   - Choose \( t = 1 \):
     - Velocity: \( v(1) = 3(1)^2 - 12(1) - 96 = 3 - 12 - 96 = -105 \) (negative).
     - Acceleration: \( a(1) = 6(1) - 12 = -6 \) (negative).
     - Speed is increasing because both are negative.

2. **Interval \( (2, 8) \):**
   - Choose \( t = 4 \):
     - Velocity: \( v(4) = 3(4)^2 - 12(4) - 96 = 48 - 48 - 96 = -96 \) (negative).
     - Acceleration: \( a(4) = 6(4) - 12 = 24 - 12 = 12 \) (positive).
     - Speed is not increasing because velocity and acceleration have opposite signs.

3. **Interval \( (8, \infty) \):**
   - Choose \( t = 10 \):
     - Velocity: \( v(10) = 3(10)^2 - 12(10) - 96 = 300 - 120 - 96 = 84 \) (positive).
     - Acceleration: \( a(10) = 6(10) - 12 = 60 - 12 = 48 \) (positive).
     - Speed is increasing because both are positive.

### Final Answer

The speed of the particle is increasing on the intervals:

\[
\boxed{(0, 2) \cup (8, \infty)}.
\]

Would you like more details or have any questions?

Here are five related questions that might interest you:
1. How do you find when the speed of the particle is decreasing?
2. What is the significance of acceleration turning points in motion analysis?
3. How do you determine the intervals of particle deceleration?
4. How can you find when the particle stops or changes direction?
5. What happens when acceleration is zero but velocity is not?

**Tip:** The speed increases when both the velocity and acceleration have the same sign. Always check intervals around critical points to determine their behavior!

Learn how to determine intervals of increasing speed for a particle moving along the x-axis with a step-by-step calculus solution. Explore velocity and acceleration functions to identify when the particle's speed increases. Suitable for advanced high school students and calculus enthusiasts.

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