Answer:
Show that if [tex]f(a) = f(b) = 0[/tex] for some [tex]a,\, b \in [-2,\, 2][/tex] where [tex]a \ne b[/tex], then by Rolle's Theorem [tex]f^{\prime}(x) = 0[/tex] for some [tex]x \in (-2,\, 2)[/tex]. However, no such [tex]x[/tex] exists since [tex]f^{\prime}(x) < 0[/tex] for all [tex]x \in (-2,\, 2)\![/tex].
Note that Rolle's Theorem alone does not give the exact value of the root. Neither does this theorem guarantee that a root exists in this interval.
Step-by-step explanation:
The function [tex]f(x) = x^{3} - 12\, x + 11[/tex] is continuous and differentiable over [tex][-2,\, 2][/tex]. By Rolle's Theorem. if [tex]f(a) = f(b)[/tex] for some [tex]a,\, b \in [2,\, -2][/tex] where [tex]a \ne b[/tex], then there would exist [tex]x \in (a,\, b)[/tex] such that [tex]f^{\prime}(x) = 0[/tex].
Assume by contradiction [tex]f(x)[/tex] does have more than one roots over [tex][-2,\, 2][/tex]. Let [tex]a[/tex] and [tex]b[/tex] be (two of the) roots, such that [tex]a \ne b[/tex]. Notice that [tex]f(a) = 0 = f(b)[/tex] just as Rolle's Theorem requires. Thus- by Rolle's Theorem- there would exist [tex]x \in (a,\, b)[/tex] such that [tex]f^{\prime}(x) = 0[/tex].
However, no such [tex]x \in (a,\, b)[/tex] could exist. Notice that [tex]f^{\prime}(x) = 3\, x^{2} - 12[/tex], which is a parabola opening upwards. The only zeros of [tex]f^{\prime}(x)[/tex] are [tex]x = (-2)[/tex] and [tex]x = 2[/tex].
However, neither [tex]x = (-2)[/tex] nor [tex]x = 2[/tex] are included in the open interval [tex](-2,\, 2)[/tex]. Additionally, [tex]a,\, b \in [-2,\, 2][/tex], meaning that [tex](a,\, b)[/tex] is a subset of the open interval [tex](-2,\, 2)[/tex]. Thus, neither zero would be in the subset [tex](a,\, b)[/tex]. In other words, there is no [tex]x \in (a,\, b)[/tex] such that [tex]f^{\prime}(x) = 0[/tex]. Contradiction.
Hence, [tex]f(x) = x^{3} - 12\, x + 11[/tex] has at most one root over the interval [tex][-2,\, 2][/tex].
