This is equivalent to showing that the function f(x) = x^3-15x+cf(x)=x3−15x+c cannot have two zeros in [-2,2][−2,2].
Summary:
If ff had two zeros in [-2,2][−2,2], then 3x^2-15 = 03x2−15=0 would have a solution in [-2,2][−2,2]. But 3x^2-15 = 03x2−15=0 does not have any such solution. Therefore, there ff cannot have two different zeros in [-2,2][−2,2].
Details:
Suppose that ff does have two different zeros in [-2,2][−2,2].
Call the lesser one aa and the greater, bb.
Since ff is a polynomial, ff is continuous everywhere, so, in particular, ff is continuous on [a,b][a,b].
Since ff is a polynomial, its derivative, f' is a polynomial, so f'(x) is defined for all x. Therefore, f is differentiable on (a,b).
We have supposed that f(a) = 0 and f(b) = 0
The hypotheses of Rolle's Theorem are true for this function on this interval, therefore the conclusion must also be true.
(Supposing that there are two different zeros,) there must be a c in (a,b) with f'(c) = 0
HOWEVER, f(x) = 3x^2-15 is zero only at +-sqrt5, neither of which can be in (a,b).
(Remember that a and b are both in [-2,2]. But sqrt5 > 2 and -sqrt5 < -2.)
Summary:
If f had two zeros in [-2,2], then 3x^2-15 = 0 would have a solution in [-2,2]. But 3x^2-15 = 0 does not have any such solution. Therefore, there f cannot have two different zeros in [-2,2].
