ChrisV
Find the average rate of change of the function over the given interval. Compare the average rate of change with the instant rate of change at the endpoints of the interval.
f(x)=-1/x [1,2]
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ChrisV
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f(x)=-1x^-1
-1(-1)x^-2
1/x^2
ChrisV
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1/(1)^2=1 1/(2)^2=2
instantaneous rates: f'(1)=1 ; f'(2)= 1/4
ChrisV
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how do i find the average rate?
ChrisV
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i know the average rate is 1/2 because the back of the book tells me, but I do not see how to get that from my work. :(
dumbcow
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think of slope between the 2 endpoints
avg rate = f(2) -f(1) / 2-1
= -1/2 -(-1) / 1
= 1/2
ChrisV
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so basically f(1)-f(2)?
dumbcow
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or more technically using calculus
\[avg rate = \frac{1}{b-a}\int\limits_{a}^{b}f'(x) dx = \frac{f(b) -f(a)}{b-a}\]
ChrisV
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im trying to understand really
ChrisV
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so if i take the original equation f(x)=-1/x
ChrisV
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then i do the f(2)-f(1)
ChrisV
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wait +f1)
ChrisV
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im just confusing myself now lol
dumbcow
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yeah f(2) -f(1)
ChrisV
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-1/2-(-1/1)=1/2
ChrisV
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thanks
dumbcow
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welcome