 ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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# Average Rate of Change - Problem 3

Norm Prokup ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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To find where the average rate of change of an object is positive or negative, simply find slopes between the given points (these points will be the time at which an object is at a certain position). By calculating the change in position over the change in time, this slope will be positive or negative, which will tell you whether the average rate of change is positive or negative. To find where the average rate of change is greatest, simply find between which two points the slope is the greatest. Note that positive average rates of change are greater than negative average rates of change.

It’s really important that you be able to look at the graph of a function and tell something about its average rates of change. Here is the graph of a function y equals f(x). I’m asked on which intervals is the average rate of change of f(x) positive between 0 and 10, between 10 and 30, between 30 and 40 and between 40 and 60? So let me look at those 4 intervals.

Between 0 and 10, I’m on this interval, remember that average rate of change can be measured by the slope of the secant line. This graph is linear, so the slope of the secant line is the slope of this graph. It looks positive. So definitely between 0 and 10, it’s going to be positive. Let me circle that. I should also check the other intervals.

Between 10 and 30, this interval right here, it looks like the slope is negative. So average rate of change will also be negative, so not here. Between 30 and 40, this slope is also negative, so not here. And between 40 and 60, remember that the average rate of change is the secant line from 40 to 60. So I will draw a secant line from these two. You can see that it’s going to have a negative slope, but I’ll just draw it for fun. Here’s my secant line, negative slope. So the average rate of change is not positive there.

Now let’s take a look at part B. It says compute the average rate of change of f(x) on the interval from 20 to 60. So remember the formula for average rate of change, it's f(60) minus f(20) over 60 minus 20. I need to look at my graph and find out what’s f(60) and is what f(20)? Let’s take a look.

Here’s my graph, f(60) is going to be ½ of 10, 5 and f(20) looks like it's 30. Remember that f(60) is 5, f(20) is 30. So that’s 5 minus 30 over 40. I don’t have units in this problem just y equals f(x). There are no units in the numerator, denominator. So I’m just going to calculate this value. I get -25/40 which reduces to -5/8.