To determine the interval in which the roots of the equation \(x^3 - 6x + 1 = 0\) lie, we will use the Intermediate Value Theorem and analyze the sign changes of the function \(f(x) = x^3 - 6x + 1\) over the given intervals.
The Intermediate Value Theorem states that if a continuous function \(f(x)\) changes sign over an interval \([a, b]\), then there exists at least one root in that interval.
First, evaluate the function at the boundaries of the given intervals:
- For the interval \((2,3)\):
- \(f(2) = 2^3 - 6 \cdot 2 + 1 = 8 - 12 + 1 = -3\)
- \(f(3) = 3^3 - 6 \cdot 3 + 1 = 27 - 18 + 1 = 10\)
- The sign changes from \(-3\) to \(10\), indicating a root between 2 and 3.
- For the interval \((3, 4)\):
- \(f(3) = 10\) (already calculated)
- \(f(4) = 4^3 - 6 \cdot 4 + 1 = 64 - 24 + 1 = 41\)
- The function retains a positive value, indicating no root in this interval.
- For the interval \((3, 5)\):
- \(f(3) = 10\) (already calculated)
- \(f(5) = 5^3 - 6 \cdot 5 + 1 = 125 - 30 + 1 = 96\)
- The function remains positive, hence no root in this interval.
- For the interval \((4, 6)\):
- \(f(4) = 41\) (already calculated)
- \(f(6) = 6^3 - 6 \cdot 6 + 1 = 216 - 36 + 1 = 181\)
- The function remains positive, hence no root in this interval.
From our evaluations, it is clear that the function changes sign only between \(x = 2\) and \(x = 3\), indicating a root in this interval. Thus, the correct answer is:
\((2, 3)\)