an object is moving along the x-axis such that the position of the object in masters at time t in seconds is giving by the function x(t)=t^3-12t^2+36t-5, for 0<=t<=8

a- when is the object moving right? and when to the left?

b- When is the object's velocity both positive and increasing

c- When is the object's velocity both negative and decreasing

d- You may notice the object starts 5m from the origin, but when is it furthest from the origin?

e- It can be shown that x(t)>0 when 0<=t<=8. When is the object closet to the origin?

f-Sketch a graph of x(t)

moving right ==> distance increasing

x increasing ==> dx/dt > 0

Now see what you can do with the questions. There are lots of online graphing web sites that you can use to confirm your graph ideas. (desmos, wolframalpha, ...)

To answer these questions, we need to analyze the given position function step by step. Let's go through each question:

a) To determine when the object is moving right or left, we need to consider the sign of the velocity. Velocity is the derivative of position with respect to time. Thus, we can find the velocity function, v(t), by taking the derivative of x(t). Differentiating x(t) gives us:

x'(t) = 3t^2 - 24t + 36

Now we need to determine when this velocity function is positive or negative. To do this, we can solve the inequality:

3t^2 - 24t + 36 > 0

First, let's find the critical points by setting the equation equal to zero:

3t^2 - 24t + 36 = 0

Solving this quadratic equation, we find t = 2 and t = 6.

Next, we need to determine the sign of the velocity in each of the three intervals: (0, 2), (2, 6), and (6, 8). We can pick a test point inside each interval and evaluate the velocity function.

For t = 1, v(1) = 3(1)^2 - 24(1) + 36 = 15, which is positive. Therefore, in the interval (0, 2), the object is moving to the right.
For t = 3, v(3) = 3(3)^2 - 24(3) + 36 = -9, which is negative. Therefore, in the interval (2, 6), the object is moving to the left.
For t = 7, v(7) = 3(7)^2 - 24(7) + 36 = 15, which is positive. Therefore, in the interval (6, 8), the object is moving to the right.

b) To find when the object's velocity is both positive and increasing, we need to find when the velocity function is positive and its derivative is positive. First, let's find the derivative of the velocity function:

v'(t) = 6t - 24

Now we set v'(t) greater than zero and solve:

6t - 24 > 0
6t > 24
t > 4

Hence, the object's velocity is positive and increasing for t > 4.

c) To find when the object's velocity is both negative and decreasing, we need to find when the velocity function is negative and its derivative is negative. First, let's set the velocity function v(t) less than zero:

3t^2 - 24t + 36 < 0
(t - 2)(t - 6) < 0

We analyze the sign of the velocity in the three intervals: (0, 2), (2, 6), and (6, 8), similar to part a. From the analysis, we determine that the object's velocity is negative and decreasing for t between 2 and 6.

d) The object is furthest from the origin when its position is at the maximum. We can find this maximum by finding the critical point of the position function. To do this, we take the derivative of the position function:

x'(t) = 3t^2 - 24t + 36

Setting x'(t) equal to zero:

3t^2 - 24t + 36 = 0

Solving this quadratic equation, we find t = 4. This corresponds to a maximum since the concavity of the function changes from positive to negative at t = 4. Thus, the object is furthest from the origin at t = 4 seconds.

e) The object is closest to the origin when its position is at a minimum. To find this minimum, we need to find the critical point of the position function. We already found this critical point in part d: t = 4 seconds. Therefore, the object is closest to the origin at t = 4 seconds.

f) To sketch a graph of x(t), we need to identify key features and plot the function. From the given position function, we know that x(t) = t^3 - 12t^2 + 36t - 5. We have determined important points from the previous questions:

- The object starts 5 m from the origin.
- The object is furthest from the origin at t = 4 seconds.
- The object is closest to the origin at t = 4 seconds.

Using this information, we can plot the graph of x(t) on the x-axis (time) and y-axis (position). Keep in mind the boundaries of the function (0 <= t <= 8).