Posted by **cel** on Monday, November 29, 2010 at 9:31pm.

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.

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