To determine when the particle is moving to the left, we need to analyze its velocity. The velocity is the derivative of the position function $x(t)$ with respect to time $t$.

The position function is given by:

$x(t) = t^4 - 16t^3$

Step 1: Find the velocity function

The velocity $v(t)$ is the first derivative of $x(t)$:

$v(t) = \frac{d}{dt} \left(t^4 - 16t^3 \right)$

$v(t) = 4t^3 - 48t^2$

Step 2: Set the velocity equal to zero to find critical points

We solve for $v(t) = 0$ to find when the particle changes direction:

$4t^3 - 48t^2 = 0$

Factor the equation:

$4t^2(t - 12) = 0$

Thus, the critical points are:

$t = 0 \quad \text{or} \quad t = 12$

Step 3: Determine intervals of motion

We need to analyze the sign of $v(t)$ in the intervals determined by the critical points $t = 0$ and $t = 12$. The intervals are:

1. $0 < t < 12$
2. $t > 12$

To determine the sign of $v(t)$ in these intervals, we test points within each interval.

• For $t = 1$ (in $0 < t < 12$): $v(1) = 4(1)^3 - 48(1)^2 = 4 - 48 = -44$ So, $v(t) < 0$ in $0 < t < 12$, meaning the particle is moving to the left in this interval.

• For $t = 13$ (in $t > 12$): $v(13) = 4(13)^3 - 48(13)^2 = 8788 - 8112 = 676$ So, $v(t) > 0$ in $t > 12$, meaning the particle is moving to the right in this interval.

Conclusion:

The particle is moving to the left on the interval $(0, 12)$.

Let me know if you'd like further details or explanations! Here are some related questions:

1. How can we find when the particle is moving to the right?
2. How do critical points relate to the motion of the particle?
3. What happens to the velocity as $t \to \infty$?
4. Can we determine the particle's acceleration at any given time?
5. How does the shape of the position function's graph relate to the particle's movement?

Tip: Always check the sign of the velocity when determining the direction of motion.