Continuity
Objective. Apply continuity definitions and the intermediate value theorem.
- 1. Learn
- 2. Worked example
- 3. Practice
- 4. Feedback
- 5. Continue
Learn: the key idea
A function f is continuous at a if lim(x→a) f(x)=f(a), meaning the limit exists, the function is defined there, and the two values agree; f is continuous on an interval if this holds at every point. The intermediate value theorem (IVT) states that if f is continuous on [a,b] and k lies between f(a) and f(b), then some c∈(a,b) satisfies f(c)=k, which is why continuous functions cannot "skip" values. For f(x)=x³−x−1 on [1,2], f(1)=−1 and f(2)=5, so by IVT there exists a root in (1,2). Checking all three conditions of the definition, not just that a formula exists, is essential before claiming continuity.
Worked example
Use the intermediate value theorem to show that f(x)=x³−x−1 has a root between x=1 and x=2, given continuity of f.
- 1f is a polynomial, so it is continuous on [1,2].
- 2Evaluate f(1)=1−1−1=−1.
- 3Evaluate f(2)=8−2−1=5.
- 4Since f(1)<0<f(2), by the IVT there exists c∈(1,2) with f(c)=0, a root.
Answer: A root exists in (1,2) by the IVT, since f(1)=−1 and f(2)=5
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Practice
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Start practisingCommon mistake. Students assume a function is continuous everywhere its formula is defined, overlooking points where separate pieces of a piecewise function fail to match; continuity at a junction point must be verified by comparing one-sided limits and the function value.
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