Thread: Is it normal to have two differing gradients for one point?

I have an exam tomorrow and part of it will consist of dervitives...

I was given a revision sheet and this is one of the questions:

Find the average rate of change of f(x) = x^3 - 4x^2 + x +2 from x=1 to x=4

So, I found the derivitive: f'(x) = 3x^2 + 8x + 1

Subbed in 1: f'(1) = -4
Subbed in 4: f'(4) = 17

So, average rate of change is delta y over delta x
17-(-4)/4-1
= 21/3
= 7
Therefore, average rate of change is 7...

Looked at the revision sheet answer and the answer was 2.
I realise how they've gotten it... Just subbed in the values into the original equation.

What I don't get is how there's two different averages. Am I doing something wrong? -It's a test on derivitives so it defeats the purpose if you don't work the derivitive out... Hmmm.

Which method should I use??
Help would be appreciated.

Originally Posted by Lucille

Find the average rate of change of f(x) = x^3 - 4x^2 + x +2 from x=1 to x=4

So, I found the derivitive: f'(x) = 3x^2 + 8x + 1

You messed up here.

Can you please tell me how I messed it up?

Do you mean with the derivitive?

f'(x) = 3x^2 + 8x + 1

--- I'm sorry, it was a typing error. My working out says it's"

f'(x) = 3x^2 - 8x + 1

Originally Posted by Lucille

Can you please tell me how I messed it up?

the derivative gives the instantaneous rate of change, we do not need f'(x) here.

for a function , the average rate of change between and is given by:

......that is, the slope of the secant line connecting the two points

Thankyou so much. I was wondering why it wasn't working. That makes sense.

I just wondered why they would put something like that in a derivitive test.

Originally Posted by Lucille

Thankyou so much. I was wondering why it wasn't working. That makes sense.

I just wondered why they would put something like that in a derivitive test.

well, it is the basic structure from which the derivative evolved. remember, the derivative is actually the limit as a gets close to b of the secant line

that is,

or equivalently,

this may help