# Problem verification

#### CleMatt

##### New member
Hello, I would like some help verifying i have the correct answers to make sure im on the right track. Any help would be greatly appreciated thank you!

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#### Dr.Peterson

##### Elite Member
Let's at least make the problem easier to see:

#### HallsofIvy

##### Elite Member
$$\displaystyle s(t)= 3t^2- t^3= t^2(3- t)$$. $$\displaystyle t^2$$ is non-negative for all t and 0 only for t= 0, 3- t= -(t- 3) is positive for t< 3, 0 at t= 3, and negative or t>3. So for t< 0, s(t) is the product of two positive numbers so is positive. s(t)= 0. s(t) is the product of two positive number so is positive again for 0< t< 3. s(3)= 0. s(t) is the product of positive and a negative so is negative for t> 3.

So, the graph, as we go from negative to positive, comes down from the top left corner of the graph ("from infinity) to (0, 0) where it is tangent to the s-axis, goes back up to some maximum value, then back down to (3, 0) and then down to the lower right corner of the graph (to negative infinity.

Do you understand that the velocity function and acceleration function are the first and second derivatives of $$\displaystyle s(t)= 3t^2- t^3$$? Do you know what those derivatives are?

#### CleMatt

##### New member
This was how i approached the problem, i just don't think i am on the right track, so any reviews and feedback would be great!

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• 523 KB Views: 2

#### CleMatt

##### New member
This was how i approached the problem, i just don't think i am on the right track, so any reviews and feedback would be great!
okay, i think i have just noticed my data entry on my calculator has completely thrown me for my graphs

#### CleMatt

##### New member

I believe these graphs are an actual representation of the data.

Although for the third question, are they asking to supply the answers for both graphs, or to get one single answer from utilizing both graphs?

#### HallsofIvy

##### Elite Member
It is impossible to tell what they are asking for because you have not told us what the question is! Apparently you are given the function s(t)= 3t^2- t^3 and are asked to graph it and its derivative, s'(t)= 6t- 3t^2. Is that right? If so then the graphs in your second post are correct.

For III you calculate the derivative of $$\displaystyle f(x)= 3x^2$$ at x= 2 using the limit of the difference quotient. Is that what the problem asked you to do? Yes, that derivative is 6(2)= 12.