Wednesday

January 18, 2017
Posted by **bill** on Monday, February 28, 2011 at 10:07pm.

- calc -
**MathMate**, Tuesday, March 1, 2011 at 9:51amHint:

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). - calc -
**Peter**, Saturday, December 3, 2016 at 10:38pmThank 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.