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
Answered by Paul T. Maths tutor

11467 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

The function f is defined by f(x)= 2/(x-3) + x - 6 . Determine the coordinates of the points where the graph of f intersects the coordinate axes.


Differentiate f = ln(x^2 + 1) / (x ^ 2 + 1).


A curve is defined by the parametric equations x=t^2/2 +1, y=4/t -1. Find the gradient of the curve when t =2.


Draw the curve for x^2-5x+6


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2025 by IXL Learning