Precalculus -- determine the rate of change of the function

Niaboc67
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Homework Statement



BsGt5xN.png


Homework Equations


Rate of change formula: y2-y1/x2-y2

The Attempt at a Solution


If this is a rate of change problem so I use the rate of change formula: y2-y1/x2-y2
and since it's from 2 to 4 I think that would be (2,0) (4,0)
and thus (0-0)/(4-2) = 0/2 = 0

I don't think this is right. What am I doing wrong?
 
on Phys.org
Niaboc67 said:

Homework Statement



BsGt5xN.png


Homework Equations


Rate of change formula: y2-y1/x2-y2

The Attempt at a Solution


If this is a rate of change problem so I use the rate of change formula: y2-y1/x2-y2
and since it's from 2 to 4 I think that would be (2,0) (4,0)
and thus (0-0)/(4-2) = 0/2 = 0

I don't think this is right. What am I doing wrong?
Your answer is correct.
Your formula needs parentheses, though. y2 - y1/x2 - x1 means ##y_2 - \frac{y_1}{x_2} - x_1##.
 
Yes, your are correct. Since it is a curved function and we are taking the rate of change on that curved interval, it will give us the average rate of change. If you start at zero, and end at zero, on average your rate of change is 0.
 
If you start at anything and end at the same thing the averge rate of change is zero, but they have given you a simple example. :D
 

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