#1

Before you all flip out, i know there is a math thread but i dont have the time to wait around for someone to stumble upon it.

the question is this:

Find the average rate of change of the function over each interval.

f(x)=x^3+1 and the two intervals are [2,3] and [-1,1]

If anyone knows how to do this please show your steps so i can do the rest of my homework on my own. I appreciate it guys.

the question is this:

Find the average rate of change of the function over each interval.

f(x)=x^3+1 and the two intervals are [2,3] and [-1,1]

If anyone knows how to do this please show your steps so i can do the rest of my homework on my own. I appreciate it guys.

#2

Say, "Only God can solve this problem." It's really hilarious what teachers say to that.

#3

ROC is the second derivative I think, Don't trust me, I'm drunk. So differentiate your function, and then stick in the limits.

#4

Maths thread

#5

The average rate of change, I believe is just gonna be f(b)-f(a)/(b-a) with interval [a,b].

#6

Just write the answer: X= -4, and f= cos6

Oh change in intervals. That should end up as f(a)-f(b)/(b-a). The interval is [a,b].

Oh change in intervals. That should end up as f(a)-f(b)/(b-a). The interval is [a,b].

*Last edited by Fryer Mike at Sep 17, 2008,*

#7

ROC is the second derivative I think,Don't trust me, I'm drunk.So differentiate your function, and then stick in the limits.

LOLOLOL!

To find the rate of change of a function at a point, take its derivative and plug in the coordinates at the point you want.

And for average rate of change, it's (f(a) - f(b)) / (a - b).

#8

LOLOLOL!

To find the rate of change of a function at a point, take its derivative and plug in the coordinates at the point you want.

Seriously, man. Confirm my shaky grasp of maths? Also, do you find after about two pints, you can do maths really, really clearly, but after about four, you're completely out of it? Also, Don't mix Cider and black Sambuca.

#9

can any of you guys like work it out because ive got like six of these to do and if i had the steps i could probably do it.

#10

can any of you guys like work it out because ive got like six of these to do and if i had the steps i could probably do it.

Goooooogle?

Wikipedia.

Maybe McKellen will do it for you, if you're lucky.

#11

Maybe McKellen will do it for you, if you're lucky.

Nigga I got kids to feed.

#12

Nigga I got kids to feed.

Nigga plz, you ain't got no kids.

#13

I hate calculus, I suck so bad at it. I skipped it in my preliminary exam for year 11.

#14

Before you all flip out, i know there is a math thread but i dont have the time to wait around for someone to stumble upon it.

the question is this:

Find the average rate of change of the function over each interval.

f(x)=x^3+1 and the two intervals are [2,3] and [-1,1]

If anyone knows how to do this please show your steps so i can do the rest of my homework on my own. I appreciate it guys.

f(x)=x^3+1

interval: [2,3]

f(2)=9

f(3)=28

average rate of change

(28-9)/(3-2) =

**19**

interval: [-1,1]

f(-1)=0

f(1)=2

average rate of change

(2-0)/(1- (-1)=

**1**

#15

http://www.algebralab.org/studyaids/studyaid.aspx?file=Calculus_6-22.xml

Thur.

Thur.

Nigga plz, you ain't got no kids.

#16

Before you all flip out, i know there is a math thread but i dont have the time to wait around for someone to stumble upon it.

the question is this:

Find the average rate of change of the function over each interval.

f(x)=x^3+1 and the two intervals are [2,3] and [-1,1]

If anyone knows how to do this please show your steps so i can do the rest of my homework on my own. I appreciate it guys.

7. Its easy. The quantum chalaka gives you the faloopah conversion number thus giving you 6.732982873812773429801778473724. Round it off.

#17

Somehow, I read the original post as rate of rate of change. Which I thought was odd. Should've payed moar attention.

*sways*

*Falls of chair*

*sways*

*Falls of chair*

#18

Errr, firstly this is average rate of change so I'm not sure that this on it's own really qualifies as 'calculus', but meh.

What everyone has been suggesting is that you use the formula f(b)-f(a)/(b-a), where a and b are your limits over which you want to find the average rate of change. It's more important you understand what's going on than just know the steps: That formula just calculates how far the function rises or falls along the y-axis on the top half, and then divides it by how far along it runs along the x-axis. (you must have been taught that gradient is rise/run or y/x, also gradient is average rate of change)

So, the steps would be:

1. plug the lower interval into your function f(x), calculate what it is, write down f(a) = (whatever the answer to that is).

2. repeat for the upper interval, write down f(b) = (whatever that was)

3. work out b-a (that is your upper interval minus your lower interval) and write that down.

4. now, take all those answers you've written down and put them into the formula f(b)-f(a)/(b-a).

I think that's right, cba to check for typos or mistakes.

What everyone has been suggesting is that you use the formula f(b)-f(a)/(b-a), where a and b are your limits over which you want to find the average rate of change. It's more important you understand what's going on than just know the steps: That formula just calculates how far the function rises or falls along the y-axis on the top half, and then divides it by how far along it runs along the x-axis. (you must have been taught that gradient is rise/run or y/x, also gradient is average rate of change)

So, the steps would be:

1. plug the lower interval into your function f(x), calculate what it is, write down f(a) = (whatever the answer to that is).

2. repeat for the upper interval, write down f(b) = (whatever that was)

3. work out b-a (that is your upper interval minus your lower interval) and write that down.

4. now, take all those answers you've written down and put them into the formula f(b)-f(a)/(b-a).

I think that's right, cba to check for typos or mistakes.

#19

the rate of change is apparent in the first derivative of the function

in your case

f(x)=x^3+1

f'(x) = 3 * x^2

the find the average value of function g(x)

(1/(a-b)) * (integral from b to a of g(x)*dx)

probably making this more complicated than this has to be

in your case

f(x)=x^3+1

f'(x) = 3 * x^2

the find the average value of function g(x)

(1/(a-b)) * (integral from b to a of g(x)*dx)

probably making this more complicated than this has to be

*Last edited by seljer at Sep 17, 2008,*

#20

(x)=x^3+1 Pft... Just divide by zero and OH SHI-

#21

f(x)=x^3+1

interval: [2,3]

f(2)=9

f(3)=28

average rate of change

(28-9)/(3-2) =19

interval: [-1,1]

f(-1)=0

f(1)=2

average rate of change

(2-0)/(1- (-1)=1

This is correct, but please take out a piece of paper, draw a graph, draw the intervals on it and any line you like between it to represent the function (it really doesn't matter if you know what the function looks like, just do it, now, seriously). Now, think to yourself, what will be the y-coordinate of this squiggly line where it crosses the two boundaries. The answer will be whatever number you get out of the function when you substitute x for the x-coordinate. Right, so you now know the coordinates of the two points, just calculate the gradient between them, which gives you the average rate of change.

When you get to the exam, knowing how to do that because you know what's going on will be a lot easier than trying to cram all the steps into your memory.

#22

the rate of change is apparent in the first derivative of the function

in your case

f(x)=x^3+1

f'(x) = 3 * x^2

the find the average value of function g(x)

(1/(a-b)) * (integral from b to a of g(x)*dx)

probably making this more complicated than this has to be

That was pretty much my original thought because it was titled 'Calculus problem'. Just to get some differentiation and integration in there.

Of course solving the integral of g(x) from b to a leaves f(b) - f(a) / (b-a) anyway. I wonder if this is how it's derived formally....

ps: what happened to your minus sign (a-b -> b-a)? It's been a over a year since I did any maths.......

#23

I think for average rate of change you integrate the function over the interval and divide by the length of the interval. I haven't done calculus in two years so if I'm wrong, then...whoops.