Absolute MAX and MIN values - Calculus I

[qwertyuiop]

I am having trouble with this problem and wanted to know if someone could help me. Now i know that velocity is the first derivative of this equation.
given the following equation:

Find the ABSOLUTE MIN and MAX values of the velocity for 't' on the closed interval [1,5]

s(t) = t^4 - 8t^3 + 18t^2 + 60t - 8

s(t) = 4t^3 - 24t^2 + 36t + 60********I believe i have started correctly but am not sure.......this should be the first derivative. from here I do not know how to find the ABSOLUTE MIN and MAX values for for 't' on the closed interval [1,5].......do I have to equal the first derivative to '0'? and if so isnt this process only done to the second derivative (acceleration)? Im trying to understand the process. I hope my questions make sense it's just a little difficult to explain something that I do not fully understand. Thank you for your help.

 

[stapel]

Find the ABSOLUTE MIN and MAX values of the velocity for 't' on the closed interval [1,5]

s(t) = t^4 - 8t^3 + 18t^2 + 60t - 8

Is s(t) the position function?

s(t) = 4t^3 - 24t^2 + 36t + 60

Is this meant to be the derivative, s'(t)? If so, then this is correct.

I do not know how to find the ABSOLUTE MIN and MAX values for for 't' on the closed interval [1,5].......do I have to equal the first derivative to '0'? and if so isnt this process only done to the second derivative (acceleration)?

The only difference between finding the "regular" max/min points and the "absolute" max/min points is the fact that you're finding the max/min points on a restricted interval.

For instance, if you were finding the max/min point of a quadratic, overall, then you'd be finding the vertex. But if you were finding the max/min point of that quadratic on a restricted interval, and especially if that interval did not include the vertex, then you'd be finding the values at the endpoints of that interval.

For further info, please try here.

 

[qwertyuiop]

Help with ABSOLUTE MIN and MAX values - Calculus I

I have the following problem:

Find the absolute min and max values for the velocity of 't' on a closed interval [1,5]

s(t) = t^4 - 8t^3 + 18t^2 + 60t - 8

s(t) = 4t^3 - 24t^2 + 36t + 60******now i know that velocity is the first derivative but from here I do not know how to find the absolute max and min values

of a closed interval [1,5]. Do I have to equal the equation to '0'? and if so, I thought that process was only done on the second derivative (acceleration). sorry I hope i dont confuse with what im writing its just difficult for me to ask about something i do not fully understand. Thank you for your help.

 

[Deleted member 4993]

I have the following problem:

Find the absolute min and max values for the velocity of 't' on a closed interval [1,5]

s(t) = t^4 - 8t^3 + 18t^2 + 60t - 8

s(t) = 4t^3 - 24t^2 + 36t + 60******now i know that velocity is the first derivative but from here I do not know how to find the absolute max and min values

of a closed interval [1,5]. Do I have to equal the equation to '0'? and if so, I thought that process was only done on the second derivative (acceleration). sorry I hope i dont confuse with what im writing its just difficult for me to ask about something i do not fully understand. Thank you for your help.

I think the absolute min/max is synonymous to GLOBAL min/max.

In that look at the value of the function [v(t)] at the end-points of the domain, in addition to the points where v'(t) =0.

If you plot v(t), this will be lot clearer.

 

[stapel]

Find the absolute min and max values for the velocity of 't' on a closed interval [1,5]

s(t) = t^4 - 8t^3 + 18t^2 + 60t - 8

s(t) = 4t^3 - 24t^2 + 36t + 60

Is this supposed to be s'(t)? If so, then this is correct.

I do not know how to find the absolute max and min values of a closed interval [1,5].

The only difference between finding "regular" max/min points and finding "absolute" max/min points is that, in the latter case, you have a restricted interval so you have to look at the interval endpoints in addition to the critical points.

For instance, if you're trying to find the overall max/min point of a parabola, you know that it's going to be at the vertex; considering the entire parabola, the max/min is at the vertex. However, if you've been given a restricted interval, especially if that interval does not vertex, then the max/min points will be at the interval endpoints. In particular, imagine working with an upward-opening parabola. The graph, as a whole, will have a min at the vertex, but will have no max. On the other hand, if you are restricted to an interval containing just a portion of the right-hand "half" of the graph, then the min will be at the left-hand interval endpoint, and there will be a max point (at the right-hand interval endpoint).

So, to answer this particular question, find the critical points, and then also consider the endpoints.

 

[Deleted member 4993]

I have the following problem:

Find the absolute min and max values for the velocity of 't' on a closed interval [1,5]

s(t) = t^4 - 8t^3 + 18t^2 + 60t - 8

s'(t) = v(t) = 4t^3 - 24t^2 + 36t + 60******now i know that velocity is the first derivative but from here I do not know how to find the absolute max and min values

of a closed interval [1,5]. Do I have to equal the equation to '0'? and if so, I thought that process was only done on the second derivative (acceleration). sorry I hope i dont confuse with what im writing its just difficult for me to ask about something i do not fully understand. Thank you for your help.

You need to find absolute max/min of v(t).

So you have differentiate it one more time and set v'(t) = 0. That will give you the local max/min points. For GLOBAL max/min points, you will need to check the values at end-points of the domain also. In this case, those would be x=1 and x=5.

 

[qwertyuiop]

am sorry yes...i wrote it wrong it is supposed to be s'(t)


s'(t) = v(t) = 4t^3 - 24t^2 + 36t + 60

s''(t) = a(t) = 12t^2 - 48t + 36

a(t) = 0

12t^2 - 48t + 36 = 0
12 (t - 1)(t - 3) = 0
*******************im not sure if this is correct but if it is what do i do with the 12?
t = 1?
t = 3?
are these the values that i have to plug into the second prime of s" to find the min and max values?

 

[stapel]

s'(t) = v(t) = 4t^3 - 24t^2 + 36t + 60

Unless you're required to work with velocity, etc, it's usually better to stick with just the derivative, s'(t), rather than switching to v(t).

s''(t) = a(t) = 12t^2 - 48t + 36

Same comment here: just use s"(t) = 12t^2 - 48t + 36

12t^2 - 48t + 36 = 0
12 (t - 1)(t - 3) = 0
*******************im not sure if this is correct but if it is what do i do with the 12?

Is twelve ever equal to zero? No? So then it does not provide any critical-point info, and also it may legally be divided off. (This is like back in algebra when you were solving quadratic equations by factoring.)

t = 1?
t = 3?
are these the values that i have to plug into the second prime of s" to find the min and max values?

What have you learned about the relationship between the zeroes of the second derivative and the shape of the original function?