Solve without using the mean value theorem: suppose that 3< = f'(t) < = 5 for all values of t. think of f(t) as the position of a moving point and f'(t) as the velocity. what are the maximum and minimum possible vales of f(8) -f(3)? explain how you got your answer.

Oh wow. My first Calculus problem I've seen on here.
First off, let me refer you to a chart my Calculus Teacher showed us. It's quite helpful when discerning what to do with these types of problems. 

3 ≤ f'(t) ≤ 5

Remember, f'(t) is the same as the slope of the derivative (So, the y value). For all values of t. Tells us that the derivative's slope will never be less than 3 or greater than 5. That means that ALL of our values of t will be above the x axis. That means our original graph will be ALWAYS increasing.

 f(8) assuming we start at 0 position, if we plug in 8 to our velocity function we get: 

3 ≤ f'(8) ≤ 5

Remember, velocity equals the distance divided by time (t). So at max we can travel 5 units of distance per 1 unit of time. 

So our maximum value of that is 8*5 = 40
Our minimum value of that is 8*3 = 24

(We can do this because our distance is equal to velocity multiplied by time) 
(Don't forget the negative!)
-f(3) has  a maximum value of: 3*5 = -15 
-f(3) has a minimum value of: 3*3 = -9

Take the difference of each of those to find which one is the maximum values.

40 - 15 = 25
40 - 9 = 31
24 - 15 = 9
24 - 9 = 15

Our maximum value (if this problem was meant to be a difference problem, if not then use the values above) is 31. 

Our minimum value (if this problem was meant to be a difference problem, if not then use the values above) is 9.

If you need help with reading the chart, then just ask. Also if you need further explanation or anything like that then ask. 

A. p = 2(l + w)p/2 - l = ww = p/2 - l b. while isolating the variable w in the equation, i first divided both sides by 2 and then i subtracted l from both sides. i hope this .

step-by-step explanation:

only stephen hawking would know.