This is equivalent to showing that the function #f(x) = x^3-15x+c# cannot have two zeros in #[-2,2]#.
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]#.
Details:
Suppose that #f# does have two different zeros in #[-2,2]#.
Call the lesser one #a# and the greater, #b#.
Since #f# is a polynomial, #f# is continuous everywhere, so, in particular, #f# is continuous on #[a,b]#.
Since #f# 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]#.
