Finding Roots Using IVT

Real Analysis · Axiom Academy

EXAMPLE Finding Roots Using the IVT Use the Intermediate Value Theorem to prove x^3 - x - 1 = 0 has a root in [1, 2] , then bisect to pin it down. Show that the equation x^3 - x - 1 = 0 has a solution in the interval [1, 2] using the Intermediate Value Theorem, then use one round of bisection to trap the root inside a smaller interval. The picture: one sign change, one crossing The graph of y = x^3 - x - 1 sits below the axis at x = 1 (where y = -1 ) and above it at x = 2 (where y = 5 ). A continuous curve can't jump the gap without touching the axis — so it crosses y = 0 somewhere in between. Bisection then halves the bracket each round. The sign change f(1) < 0 < f(2) is what forces a crossing. Bisection keeps the half where the sign still flips, shrinking the bracket toward . Nice work — you proved a root exists with the IVT, then bisection turned "somewhere in [1,2] " into a tight bracket. The IVT needs continuity: always check f is continuous on the closed interval [a,b] first. Polynomials are continuous everywhere, so that step is automatic here. Opposite signs force a root: if , then f goes from below the axis to above (or vice versa), so it must cross zero in between. Existence, not location: the IVT guarantees some c with f(c) = 0 , but doesn't say where — it only promises the crossing exists. Bisection pins it down: check the midpoint's sign and keep the half where the sign still changes. Each round halves the interval — here .

This is the written version of the interactive lesson above. See the full Real Analysis course.