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Answer:
Since the function [tex]f(x) = x^{3} - x^{2} - 5[/tex] is continuous, [tex]f(2) < 0[/tex], and [tex]f(3) > 0[/tex], by the intermediate value theorem, there must exist a real number [tex]2 < x < 3[/tex] for which [tex]f(x) = 0[/tex].
Step-by-step explanation:
By the Intermediate Value Theorem, if a function [tex]f[/tex] is continuous, and that [tex]f(a) < f(b)[/tex] for real numbers [tex]a[/tex] and [tex]b[/tex] (both are constants,) then for any given function value [tex]y[/tex] where [tex]f(a) < y < f(b)[/tex], there would exist one or more [tex]x[/tex] where [tex]a < x < b[/tex] such that [tex]f(x) = y[/tex].
In this question, let [tex]f(x) = x^{3} - x^{2} - 5[/tex]. This function would be continuous as it is a polynomial.
The goal is to show that there exists a solution to the equation [tex]x^{3} - x^{2} - 5 = 0[/tex] (or, equivalently, [tex]f(x) = 0[/tex]) in the given interval [tex]2 < x < 3[/tex]. To prove this statement with the help of the Intermediate Value Theorem, set up the premise as follows:
Since [tex]y = 0[/tex] is between [tex]f(a) = (-1)[/tex] and [tex]f(b) = 13[/tex], it is indeed true that [tex]f(a) < y < f(b)[/tex]. Therefore, by the Intermediate Value Theorem, there would exist some [tex]x[/tex] where [tex]a < x < b[/tex] (in other words, [tex]2 < x < 3[/tex]) such that [tex]f(x) = 0[/tex]. In other words, there would exist at least one solution in the interval [tex](2,\, 3)[/tex] to the equation [tex]x^{3} - x^{2} - 5 = 0[/tex].