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