Average rate of change question

Scorpy

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Jan 27, 2012
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The average rate of change is the slope of the secant line from [a,b]

Whereas, the instantaneous rate of change is the slope at said point.


Average:

21_3daa5e80c14c185d68267a69a969f0cf.png


Instantaneous:

60_5da79965f04068e3634411bb82aa49ab.png


Find
20_48d097455b7e6e4e063486607b3a5b06.png


25_793fde039320a1b804aaae01eb898f7e.png


68_6ad9e3b48a863c964b5dc717d32fe735.png





I just wanted to ask if you want to find the average rate of change over some interval, and what if the slope tends to infinity? What next?

Thanks
 
The average rate of change is the slope of the secant line from [a,b]

Whereas, the instantaneous rate of change is the slope at said point.


Average:

21_3daa5e80c14c185d68267a69a969f0cf.png


Instantaneous:

60_5da79965f04068e3634411bb82aa49ab.png


Find
20_48d097455b7e6e4e063486607b3a5b06.png


25_793fde039320a1b804aaae01eb898f7e.png


68_6ad9e3b48a863c964b5dc717d32fe735.png





I just wanted to ask if you want to find the average rate of change over some interval, and what if the slope tends to infinity? What next?

Thanks

Instantaneous rate:

\(\displaystyle \lim_{h\to 0}\dfrac{f(x_1+h)-f(x_1)}{h}\)

\(\displaystyle = \ \lim_{h\to 0}\dfrac{\dfrac{-6}{x_1+h}-\dfrac{-6}{x_1}}{h}\)

\(\displaystyle = \ 6*\lim_{h\to 0}\dfrac{h}{h*(x_1)(x_1+h)}\)

\(\displaystyle = \ \dfrac{6}{{x_1}^2}\)
 
I'm sorry about the double post :)
And I think I did not make my question right. I was asking what would be the tangent line over interval [a,b] (y-b=m(x-a)) if the slope e.g the limit tends to infinity?
 
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