The equation x^3 - 3*x + 1 = 0 has three real roots; Show that one of the roots lies between −2 and −1
In order to prove that one real root of an equation is situated in a certain interval, we calculate the value of the function at the ends of the given interval. In the given case, f(-2) = (-2)^3 - 3*(-2) + 1 = -1 and f(-1) = (-1)^3 - 3*(-1) + 1 = 3. As our function is an elementary one (a polynomial), it is continuous over all real values, which means that the function will take all real values from -1 to 3 as x goes from -2 to -1, including 0. This means that one of the roots of f lies in the interval (-2, -1).
PT
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