An object moves along the x -axis, its position at each time...

An object moves along the $x$ -axis, its position at each time $t \geq 0$ given by $x(t) .$ Determine the time interval(s), if any, during which the object satisfies the given condition.
$$x(t)=t^{3}-12 t^{2}+21 t ; \text { moves left }$$

Key Concepts

Derivative and Its Sign

The derivative of a function is a fundamental tool in calculus that provides information about the function’s increasing or decreasing behavior. For motion analysis, the sign of the derivative (velocity) tells us the direction of travel: a negative derivative indicates that the object is moving in the opposite direction (for instance, moving left on the x-axis).

Critical Points

Critical points occur where the derivative of the function is zero or undefined. These points are crucial because they help in partitioning the time domain into intervals where the function—and hence, the motion—changes behavior. Analyzing these points and the sign of the derivative in the resulting intervals helps determine when the object moves in a particular direction.

Velocity

Velocity represents the rate of change of the object’s position with respect to time, and is obtained by differentiating the position function. It indicates both the speed and the direction of the motion along a given axis.

Transcript

00:01 So when we want to find where our object is moving to the left, that means we want to find where is our velocity negative.

00:07 And so we want to find a velocity equation, v of t, and in this case it's equal to the derivative of our displacement equation, x of t.

00:15 And so here it would be equal to the derivative of t to the third, which is 3t squared.

00:20 I'm just using the power rule to find that derivative, and then minus 12 times the derivative of t squared, which would be 24t, since the derivative of t squared is 2t.

00:29 And then plus 21 times the derivative of t and the derivative of t is equal to 1 so that's just going to be 21.

00:36 And so we want to figure out where is this function equal to 0.

00:41 So let's go ahead and set it equal to 0.

00:43 And i'm going to factor out a 3 on this right side.

00:45 So we have 3 times t squared minus 8t plus 7.

00:51 And so this actually factors into t minus 7 times t minus 1.

00:57 So that means at t is equal to 7 and at t is equal to 1 is where our velocity is going to be equal to 0.

01:04 And we want to know where velocity is equal to zero because in order for our velocity to change sign, it needs to have a value of zero at some point, since this is a continuous function for it to go from negative to positive, it has to cross the x -axis, which means we have to have a value of v of t equal to zero.

01:24 And so we're really looking at the values before, so values before t or t is equal to one, so values t less than one, and then values between one and seven, and then values of t that are greater than seven.

01:39 And the reason for this is because, again, in order to change sign, we have to cross the x axis.

01:45 These are the only two places where we would potentially cross the x axis.

01:49 We could also just touch it.

01:51 And by that i mean just go up, touch the x axis.

01:54 Axis.

01:55 So if we were below the x -axis, we would go up, touch it, and then go back to being below the x -axis, or we could actually cross it...

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