# Thread: Average Velocity and Functions

1. ## Average Velocity and Functions

First of all, would this work for average velocity for this:

The height of Heather's calculus book t seconds after she drops it from the top of the 500-meter-tall building is s(t)=-4.9t^2+500.

a. Find the average velocity of the object during the first 3 seconds.

b. Use the Mean Value Theorem to verify that at some time during the first 3 seconds of fall the instantaneous velocity equals the average velocity. Find that time.

s(t)=-4.9t^2+500,
-4.9(3)^2+500=455.9,
500-455.9=44.1

a. Vav= displacement/time=44.1/3=14.7?

and then I don't get B...

Also...

Make a function having all the indicated characteristics:

f(0)=f(6)=0
f '(3)=f '(5)=0
f '(x)>0 if x<3
f '(x)>0 if 3<x<5
f ''(x)<0 if x<3 and x>4
f ''(x)>0 if 3<x<4

2. Originally Posted by Th3On3Fr33man
First of all, would this work for average velocity for this:

The height of Heather's calculus book t seconds after she drops it from the top of the 500-meter-tall building is s(t)=-4.9t^2+500.

a. Find the average velocity of the object during the first 3 seconds.

b. Use the Mean Value Theorem to verify that at some time during the first 3 seconds of fall the instantaneous velocity equals the average velocity. Find that time.

s(t)=-4.9t^2+500,
-4.9(3)^2+500=455.9,
500-455.9=44.1

a. Vav= displacement/time=44.1/3=14.7?

and then I don't get B...
you simply want to solve s'(t) = average velocity, for t

Also...

Make a function having all the indicated characteristics:

f(0)=f(6)=0
f '(3)=f '(5)=0
f '(x)>0 if x<3
f '(x)>0 if 3<x<5
f ''(x)<0 if x<3 and x>4
f ''(x)>0 if 3<x<4
check these characteristics again, make sure you didn't make a typo

3. no typo, thats it

4. Originally Posted by Th3On3Fr33man
no typo, thats it
you only told us where the function is increasing, but you did not mention where it is decreasing (as we know it has to to hit the two zeros). you must have some information on where f'(x) < 0

5. Ok, sorry, didn't see it before; add f'(x)<0 if x>5

6. Originally Posted by Th3On3Fr33man
Ok, sorry, didn't see it before; add f'(x)<0 if x>5
it seems to me this will look like a quartic function, but i think we also need something like f'(x)<0 for 3< x < 4 or something. otherwise the relative extrema at x = 3 cannot be a max or min, it would be some weird inflection point

7. Ok I'll retype it exactly as it says:

Sketch a graph of a function having all the indicated characteristics:

f (0) = f (6) = 0
f ' (3) = f ' (5) = 0
f ' (x) > 0 if x < 3
f ' (x) > 0 if 3 < x < 5
f ' (x) < 0 if x > 5
f " (x) < 0 if x < 3 and x > 4
f " (x) > 0 if 3 < x < 4

That should be exactly what it said