--- title: "Locating Roots and Iterative Methods" description: "Trap a root between two values by a change of sign, then rearrange the equation into the form x = g(x) and iterate from a starting value until the digits stop moving." canonical: https://lightmysky.com/learn/mathematics/locating-roots-and-iterative-methods-mt_2COO6TWerM source: https://lightmysky.com/learn/mathematics/locating-roots-and-iterative-methods-mt_2COO6TWerM.md retrieved: 2026-09-12 --- > **Agent view.** This is the Markdown twin of the page, for tools and assistants. > When to use this site, and the call that answers each job: https://lightmysky.com/agent-instructions.md > API description (OpenAPI 3.1): https://lightmysky.com/openapi.json · Authentication: https://lightmysky.com/auth.md > Pricing: https://lightmysky.com/pricing.md · Catalog: https://lightmysky.com/llms.txt · Full catalog: https://lightmysky.com/llms-full.txt > Every machine-readable file on this domain: https://lightmysky.com/.well-known/ai-catalog.json > Ask for Markdown with `Accept: text/markdown`, a `.md` address, or `?mode=agent`. # Locating Roots and Iterative Methods Trap a root between two values by a change of sign, then rearrange the equation into the form x = g(x) and iterate from a starting value until the digits stop moving. Subject: Mathematics · Area: Calculus & Analysis · Ages 17 to 18 Page: https://lightmysky.com/learn/mathematics/locating-roots-and-iterative-methods-mt_2COO6TWerM ## Ready when they can - Use a sign change to show a root lies in a given interval - Rearrange an equation into an iterative formula - Run an iteration to a stated accuracy and justify stopping ## Lesson: Trapping a root, then walking in Take a continuous graph. If f(1) is negative and f(2) is positive, the curve must cross the axis somewhere between 1 and 2. That crossing is a root, and you proved it exists without solving anything. One sign change promises at least one root, though a wiggly graph can cross more than once. **Example.** For f(x) = x cubed minus x minus 3, f(1) = 1 - 1 - 3 = -3 and f(2) = 8 - 2 - 3 = 3. The signs differ, so a root sits between 1 and 2. Rewrite x cubed = x + 3 as x = the cube root of (x + 3) and feed each answer back in. From 1.5 you get 1.651, then 1.669, then 1.671, settling at 1.67 to two decimals. Rearranging is a choice, and some choices pull you in while others push you away. Picture y = g(x) against the line y = x and cobweb between them: shallow slopes spiral inward to the fixed point and steep ones spiral outward to failure. Watch the first few iterates, since converging decimals that settle digit by digit signal success. **Tip.** Never stop just because two iterates match. Matching decimals show the steps got small, not that the limit is near. For instance 2.496 is within 0.004 of 2.5, so it is correct to one decimal place, which allows an error up to 0.05. Stop with an error bound or a tight sign change interval around the value. **Recap.** Trap the root with a sign change, walk in with x = g(x), and stop only when an error bound says you are close. ## Practice 18 questions on this page, each with its working shown. ## Needs first - [Recognising Cubic, Reciprocal and Exponential Graphs](https://lightmysky.com/learn/mathematics/recognising-cubic-reciprocal-and-exponential-graphs-mt_2IJJ51rmq9) ## Opens up - [The Newton-Raphson Method](https://lightmysky.com/learn/mathematics/the-newton-raphson-method-mt_jkjJre5ETd) - [Continuity and the Intermediate Value Theorem](https://lightmysky.com/learn/mathematics/continuity-and-the-intermediate-value-theorem-mt_kaQR6cWjbV)