Show that the equation x^3 - 15x + c = o has exactly one real root.
All I know is that it has something to do with the Mean Value Theorem/Rolle's Theorem.
AP Calculus - Reiny, Monday, November 29, 2010 at 11:25pm
You must have a typo.
There are all kinds of values of c for which the above equation has 3 roots.
e.g. if c=0
x^3 - 15x = 0
x(x^2 - 15) = 0
x = 0 or x = ± √15
are you sure it wasn't x^3 + 15x + c = 0 ?
I will assume it was
It's been 50 years but if I recall, Rolle's theorem says that if there are two x-intercepts , then the derivative must be zero between those two points
so let f(x) = x^3 + 15x + c
f '(x) = 3x^2 +15 = 0
notice this is always positive, since x^2 = -5 has no real solution,
so there are no turning points, and the curve cuts only once.
You do know that every cubic must cut he x-axis at least once?
AP Calculus - Alon, Thursday, December 8, 2011 at 2:21am
The missing info it the interval [-2,2]. In this interval there could be just one root.