calc
posted by bill on .
Show that the equation x^315x+c=0 has at most one root in the interval [2,2].

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^315x+c
f'(x)=3x²15
Absolute maximum on [2,2] is when x=±2, f'(x)=415=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. 
yo, you plugged the numbers wrong bruh...
+ or  2 when you plug it into f prime of x, you get 3 bruh not 11...