Average Rate of Change
- Thread starter swolios
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pka
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- Jan 29, 2005
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Here is what you need, now you tell us why.
\(\displaystyle \int_{ - 2}^0 { - 1dx} + \int_0^3 {1dx} = ?\)
As an introductory calculus question, I would not interpret it quite the way pka did. I would just sayView attachment 3170
I am having trouble figuring this out( I am new to calculus ), I can not find an example similar to this in my book, any explanation/help would be appreciated.
average rate of change = (change of f(x)) / (change of x)
................................= [f(3) - f(-2)] / [3 - (-2)] = . . .
I am not sure how exactly to approach those integrals
So this is what I got from above: 1/5 (f(3)-f(-2))
I still do not understand where exactly the x/|x| and the 2 went from the picture of the problem I posted but at least I got somewhere, thanks for your help, if you could further elaborate/evaluate what I did, I would greatly appreciate.
f(x) is a piecewise function with a discontinuity at x=0. But the discontinuity is not relevant when you only need to evaluate the function at x=3 and at x=-2.
What is still confusing to you is the use of "absolute value" in the denominator of the function,
\(\displaystyle f(x) = \dfrac{x}{|x|} \)
If \(\displaystyle x < 0\), then the numerator is negative and the denominator is positive, so \(\displaystyle f(x) = -1\).
On the other hand, if \(\displaystyle x>0\), then both numerator and denominator are positive and \(\displaystyle f(x) = +1\).
Thus \(\displaystyle f(3) - f(-2) = 1 - (-1) = 2\), and the average rate of change = 2/5
The slope of the line from (-2,-1) to (3,1) is 2/5.
Thank you for the help, I spent most my morning learning about this, and also the professor went over this problem in class as it seemed to have 56% of the class confused as well, so with your help, and class I have figured out what you have done. Again Thanks!