Given the function $$f(x)=x^2+x$$,
find the average change (AC) in the interval $$[0, 10]$$ and $$[0,2]$$.
See development and solution
Development:
Using the definition $$$AC=\dfrac{y}{x}=\dfrac{f(a+x)-f(a)}{x}$$$ Given the interval $$[0,10]$$,
$$\Delta y=f(10)-f(0)=(10^2+10)-0=110$$
$$\Delta x=10-0=10$$
$$AC=\dfrac{\Delta y}{\Delta x}=\dfrac{110}{10}=11$$
Given the interval $$[0,2]$$
$$\Delta y=f(2)-f(0)=(2^2+2)-0=6$$
$$\Delta x=2-0=2$$
$$AC=\dfrac{\Delta y}{\Delta x}=\dfrac{6}{2}=3$$
Solution:
Interval $$[0,10]: \ AC=11$$
Interval $$[0,2]: \ AC=3$$