QUESTION IMAGE
Question
the graph of ( y = f(x) ) is given. show that the average rates of change of ( f ) over equal intervals of the domain follow a linear pattern.
( \frac{f(-6)-f(-8)}{-6 - (-8)}=\frac{6}{2}=3 )
( \frac{f(-4)-f(-6)}{-4 - (-6)}=\frac{2}{2}=1 )
( \frac{f(-2)-f(-4)}{-2 - (-4)}=\frac{-2}{2}=-1 )
( \frac{f(0)-f(-2)}{0 - (-2)}=\frac{-6}{2}=-3 )
( 3,1,-1,-3 )
did you answer the question?
Step1: Recall the formula for average rate of change
The average rate of change of a function \(y = f(x)\) over the interval \([a,b]\) is \(\frac{f(b)-f(a)}{b - a}\).
Step2: Calculate average rate of change for each interval
- For the interval \([-8,-6]\):
- From the graph, \(f(-8)=- 3\) and \(f(-6)=3\).
- Using the formula \(\frac{f(-6)-f(-8)}{-6-(-8)}=\frac{3-(-3)}{2}=\frac{6}{2} = 3\).
- For the interval \([-6,-4]\):
- From the graph, \(f(-6) = 3\) and \(f(-4)=5\).
- Using the formula \(\frac{f(-4)-f(-6)}{-4-(-6)}=\frac{5 - 3}{2}=1\).
- For the interval \([-4,-2]\):
- From the graph, \(f(-4)=5\) and \(f(-2)=3\).
- Using the formula \(\frac{f(-2)-f(-4)}{-2-(-4)}=\frac{3 - 5}{2}=-1\).
- For the interval \([-2,0]\):
- From the graph, \(f(-2)=3\) and \(f(0)=-3\).
- Using the formula \(\frac{f(0)-f(-2)}{0-(-2)}=\frac{-3 - 3}{2}=-3\).
Step3: Check the linear pattern of average - rate - of - change values
The average rate of change values are \(3,1,-1,-3\).
If we consider these values as a sequence \(a_n\) where \(n = 1,2,3,4\) (corresponding to the four intervals in order), and the common difference \(d\):
- \(a_2-a_1=1 - 3=-2\).
- \(a_3 - a_2=-1-1=-2\).
- \(a_4 - a_3=-3-(-1)=-2\).
Since the difference between consecutive average rate of change values is constant (\(d=-2\)), the average rates of change of \(f\) over equal intervals of the domain follow a linear pattern.
Snap & solve any problem in the app
Get step-by-step solutions on Sovi AI
Photo-based solutions with guided steps
Explore more problems and detailed explanations
The average rate of change values \(3,1,-1,-3\) have a common difference of \(-2\), so the average rates of change of \(f\) over equal intervals of the domain follow a linear pattern.