show that the equation x^3-15x+c=0 has at most one root on the interval [-2,2.

Answer 1

f(-2) = -8+30+c = 22+c
f(2) = 8-30+c = -22+c
df/dx = 3x^2-15 = 3(x^2-5)
df/dx=0 at x=±√5
so, f(x) has no turning points on [-2,2]

so, either f stays negative or positive on [-2,2], or it crosses somewhere. But, it can cross only once, since it does not turn around anywhere.