Answer:
Since [tex]f(0) < f(x) < f(1)[/tex], there is a solution for [tex]f(x)[/tex] between [tex]x = 0[/tex] and [tex]x = 1[/tex].
Step-by-step explanation:
A solution of the polynomial is a root such that sign of dependent value changes, either from positive to negative or from negative to positive. Let [tex]f(x) = x^{3} + 6\cdot x - 5[/tex], [tex]x[/tex] is a solution of the equation if and only if [tex]f(x) = 0[/tex]. We proceed to evaluate the function at each bound:
x = 0
[tex]f(0) = 0^{3}+6\cdot (0) - 5[/tex]
[tex]f(0) = -5[/tex]
x = 1
[tex]f(1) = 1^{3}+6\cdot (1) - 5[/tex]
[tex]f(1) = 2[/tex]
Since [tex]f(0) < f(x) < f(1)[/tex], there is a solution for [tex]f(x)[/tex] between [tex]x = 0[/tex] and [tex]x = 1[/tex].
