Hello, karamelquti79!
The winning times for the 200 meter dash can be approximated:
\(\displaystyle \;\;\;f(x)\;=\;\)-\(\displaystyle 0.05005x\,+\,146.3\), where \(\displaystyle x\) is the year.
What is the average rate of change of \(\displaystyle f\) ?
Since the function is linear, the average rate of change is the
slope: -\(\displaystyle 0.05005\)
If we are expected to use the
defintion of average rate of change,
\(\displaystyle \;\;\)it takes more work, of course.
In year \(\displaystyle a\), the winning time is: \(\displaystyle \,f(a)\;=\;\)-\(\displaystyle 0.05005a\,+\,146.3\) seconds.
\(\displaystyle h\) years later, (in the year \(\displaystyle a+h\)), the time is:
\(\displaystyle \;\;f(a\,+\,h)\;=\;\)-\(\displaystyle 0.05005(a\,+\,h)\,+\,146.3 \;= \;\)-\(\displaystyle 0.05005a\,-\,0.05005h\,+\,146.3\) seconds.
The
change in time is:
\(\displaystyle \;\;f(a\,+\,h)\,-\,f(a) \;= \;(\)-\(\displaystyle 0.05005a -0.05005h \,+\,146.3)\,-\,(\)-\(\displaystyle 0.05005a\,+\,146.3)\;=\;\)-\(\displaystyle 0.05005h\) seconds.
Over \(\displaystyle h\) years, the
average rate of change of the winning time is: \(\displaystyle \;\frac{-0.05005h}{h}\;=\;-0.05005\)
(Translation: the winning time is decreasing at the rate of 0.05005 seconds per year.)
