Consider as an example the function \(f(x)=-x^2\). At \(x=-2\) and \(x=2\), you have \(f(-2)=f(2)=-4\). The average rate of change of the function over the given interval is
\(\dfrac{f(2)-f(-2)}{2-(-2)}=\dfrac{-4-(-4)}4=0\)
so Hunter is correct.
Maggie is also correct, since \(f(x)\) has a turning point at the parabola's vertex when \(x=0\).
For a more general situation, you can invoke Rolle's theorem, which states that for a continuous function \(f(x)\) over an interval \([a,b]\) with \(f(a)=f(b)\) (which is the case for this example) that there is some \(c\) in the open interval \((a,b)\) such that \(f'(c)=0\).
Whenever \(f(a)=f(b)\), it's always true that the average rate of change will be \(0\). Provided the function isn't constant, it will always attain an extremum within the given interval, so a turning point must exist.
