question a particle moves along the x - axis so that at time t ≥ 0 its position is given by x(t)=t³ + 21t²…

Explanation:

Step1: Find the velocity function

The velocity function $v(t)$ is the derivative of the position - function $x(t)$. Using the power rule $\frac{d}{dt}(t^n)=nt^{n - 1}$, we have $v(t)=x^\prime(t)=3t^{2}+42t - 96$.

Step2: Set up the inequality

The particle is moving to the left when $v(t)<0$. So we need to solve the inequality $3t^{2}+42t - 96<0$. First, divide the entire inequality by 3 to simplify: $t^{2}+14t - 32<0$.

Step3: Factor the quadratic

Factor $t^{2}+14t - 32$ as $(t + 16)(t - 2)<0$.

Step4: Find the roots

Set $(t + 16)(t - 2)=0$. The roots are $t=-16$ and $t = 2$. Since $t\geq0$, we ignore $t=-16$.

Step5: Test intervals

We test the intervals $[0,2)$ and $(2,\infty)$. For $t\in[0,2)$, let $t = 1$. Then $v(1)=3\times1^{2}+42\times1 - 96=3 + 42-96=-51<0$. For $t\in(2,\infty)$, let $t = 3$. Then $v(3)=3\times3^{2}+42\times3 - 96=27+126 - 96=57>0$.

Answer:

The particle is moving to the left on the interval $[0,2)$.