Posted by bill on Monday, February 28, 2011 at 10:07pm.
Hint:
For f(x) to have at most one root on [-2,2], it must be strictly increasing or decreasing on that interval.
So let's examine f'(x) on the interval [-2,2].
f(x)=x^3-15x+c
f'(x)=3x²-15
Absolute maximum on [-2,2] is when x=±2, f'(x)=4-15=-11
Absolute minimum on [-2,2] is when x=0, f'(x)=-15.
Therefore f'(x) is negative on [-2,2], therefore strictly decreasing. Under these circumstances, f(x) can have at most one zero (root).
Thank you for your answer!!! My solution manual use Roll's theorem to solve it and I cannot understand it at all.
Your f'(2) and f'(-2) should equal -3 tho.