The only way that we’ll know for sure which direction the object is moving is to have the velocity in hand. Recall that if the velocity is positive the object is moving off to the right and if the velocity is negative then the object is moving to the left.

We need the derivative in order to get the velocity of the object. The derivative, and hence the velocity, is,

\[s'\left( t \right) = 6{t^2} - 42t + 60 = 6\left( {{t^2} - 7t + 10} \right) = 6\left( {t - 2} \right)\left( {t - 5} \right)\]

The reason for factoring the derivative will be apparent shortly.

Now, we need to determine where the derivative is positive and where the derivative is negative. There are several ways to do this. The method that we tend to prefer is the following.

Since polynomials are continuous we know from the Intermediate Value Theorem that if the polynomial ever changes sign then it must have first gone through zero. So, if we knew where the derivative was zero we would know the only points where the derivative *might* change sign.

We can see from the factored form of the derivative that the derivative will be zero at \(t = 2\) and \(t = 5\). Let’s graph these points on a number line.

Now, we can see that these two points divide the number line into three distinct regions. In each of these regions we **know** that the derivative will be the same sign. Recall the derivative can only change sign at the two points that are used to divide the number line up into the regions.

Therefore, all that we need to do is to check the derivative at a test point in each region and the derivative in that region will have the same sign as the test point. Here is the number line with the test points and results shown.

Here are the intervals in which the derivative is positive and negative.

\[\begin{array}{rl}{{\mbox{positive : }}}&{ - \infty < t < 2\,\,\,\,\& \,\,\,\,5 < t < \infty }\\{{\mbox{negative : }}}&{2 < t < 5}\end{array}\]

We included negative \(t\)’s here because we could even though they may not make much sense for this problem. Once we know this we also can answer the question. The object is moving to the right and left in the following intervals.

\[\begin{array}{rl}{{\mbox{moving to the right : }}}&{ - \infty < t < 2\,\,\,\,\& \,\,\,\,5 < t < \infty }\\{{\mbox{moving to the left : }}}&{2 < t < 5}\end{array}\]