# Average Rate of Change

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• Dec 23rd 2008, 09:11 PM
magentarita
Average Rate of Change
Find the average rate of change for the given function in the following interval.

F(x) = sinx + 2 [-90 degrees, 90 degrees]

My answer is 1.

Is this right or wrong?

If wrong, why?

• Dec 24th 2008, 12:23 AM
Moo
Hello,

If I understand correctly what an averate rate of change is, I'd say it's :

$\frac{F(90)-F(-90)}{90-(-90)}=\dots=\frac{3-1}{180}=\dots$
• Dec 24th 2008, 12:34 AM
samer_guirguis_2000
I think the answer is 2 if you are getting the average of the function and it is zero if you are getting the average of its derivative(rate).
The averaging of a function involves the integration of the function between the limits given divided by the interval. Of course you sould substitute the limits in radians.
This is purely mathematical.
if you graph this function you will find it a sine wave shifted upwards two units. so the average is two without computing.
• Dec 24th 2008, 12:42 AM
Chop Suey
The average rate of change, in this case, is the slope of the line containing the endpoints of the interval.

The instantaneous rate of change is the derivative, or the slope of the line tangent to a certain point.
• Dec 24th 2008, 03:21 AM
mr fantastic
Quote:

Originally Posted by samer_guirguis_2000
I think the answer is 2 if you are getting the average of the function and

it is zero if you are getting the average of its derivative(rate). Mr F says: No. The average of the derivative is NOT zero.

The averaging of a function involves the integration of the function between the limits given divided by the interval. Of course you sould substitute the limits in radians.
This is purely mathematical.
if you graph this function you will find it a sine wave shifted upwards two units. so the average is two without computing.

The question asks for the average rate of change. The correct answer has already been given by moo.

Discussion on the average of the function and average of the derivative is not helpful to the OP at the present time.
• Dec 28th 2008, 07:19 AM
magentarita
Thanks
I want to thank all who took time out to explain this question.