Answer:
Since [tex]f(x)[/tex] (a polynomial) is continuous over the interval [tex](2,\, 3)[/tex], simply showing that [tex]0[/tex] is between [tex]f(2)[/tex] and [tex]f(3)[/tex] will be sufficient to prove that [tex]f(x)[/tex] has at least one real zero over this interval.
Step-by-step explanation:
By the intermediate value theorem, if a function [tex]f[/tex] is continuous over the real interval [tex][a,\, b][/tex], then for any value [tex]y[/tex] where [tex]f(a) < y < f(b)[/tex], there exists at least one real [tex]x_{0}[/tex] in the interval [tex][a,\, b][/tex] such that [tex]f(x_{0}) = y[/tex].
Specifically, this theorem suggests that if [tex]f(x)\![/tex] is continuous over the real interval [tex][a,\, b][/tex] and [tex]f(a) < 0 < f(b)[/tex], then [tex]f(x)[/tex] would have at least one real zero [tex]x_{0}[/tex] in that interval ([tex]a < x_{0} < b[/tex] and [tex]f(x_{0}) = 0[/tex].) On a graph, [tex]f(x)[/tex] would be below the [tex]x[/tex]-axis at [tex]x = a[/tex] while above the [tex]x\![/tex]-axis at [tex]x = b[/tex]. Because [tex]f(x)[/tex] is continuous over this interval, the graph of this function must have intersected the [tex]x\!\![/tex]-axis somewhere within that interval.
Since the given function [tex]f(x)[/tex] in this question is a polynomial, this function would indeed be continuous over the real interval [tex][2,\, 3][/tex]. The intermediate value theorem would be applicable.
To prove the existence of this zero, simply show that [tex]f(2) < 0[/tex] and [tex]f(3) > 0[/tex].
[tex]f(2) = 16 + 8 - 16 - 10 - 5 = -7[/tex].
[tex]f(3) = 81 + 27 - 36 - 15 - 5 = 52[/tex].
Hence, by the intermediate value theorem, since [tex]f(x)[/tex] is continuous over the interval [tex][2,\, 3][/tex], there would be at least one real [tex]x_{0}[/tex] where [tex]2 < x_{0} < 3[/tex] and [tex]f(x_{0}) = 0[/tex].
