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

Related Maths A Level answers

All answers ▸

Differentiate x^5 + 3x^2 - 17 with respect to x


What is the binomial distribution and when should I use it?


Given a table showing grouped data and the frequency of each class, find the median Q2


The curve C has the equation y = 2x^2 -11x + 13. Find the equation of the tangent to C at the point P (2, -1).