Answer:
This does not contradict Rolle's Theorem, since f '(0) = 0, and 0 is in the interval (−64, 64).
Step-by-step explanation:
The given function is

To find
, we substitute
into the function.



To find
, we substitute
into the function.



To find
, we must first find
.

This implies that;





For this function to satisfy the Rolle's Theorem;
It must be continuous on
.
It must be differentiable on
.
and
.
All the hypotheses are met, hence this does not contradict Rolle's Theorem, since f '(0) = 0, and 0 is in the interval (−64, 64) is the correct choice.