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

• Sep 15th 2007, 08:08 PM
Lucille
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.
• Sep 15th 2007, 08:11 PM
Krizalid
Quote:

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.
• Sep 15th 2007, 08:18 PM
Lucille
Can you please tell me how I messed it up?
• Sep 15th 2007, 08:20 PM
Lucille
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
• Sep 15th 2007, 08:22 PM
Jhevon
Quote:

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 $f(x)$, the average rate of change between $x = a$ and $x = b$ is given by:

$\mbox {Average Rate of Change} = \frac {f(b) - f(a)}{b - a}$ ......that is, the slope of the secant line connecting the two points
• Sep 15th 2007, 08:27 PM
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.
• Sep 15th 2007, 08:36 PM
Jhevon
Quote:

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, $f'(a) = \lim_{x \to a} \frac {f(x) - f(a)}{x - a}$

or equivalently, $f'(x) = \lim_{h \to 0} \frac {f(x + h) - f(x)}{h}$

this may help