MATH1851 Chap.4 Continuity and the Intermediate Value Theorem
Continuity and the Intermediate Value Theorem
Continuity is a three-condition test
At an interior point, continuity requires a defined function value, an existing two-sided limit and equality between the two. The order matters because each condition answers a different question.
A graph can have a perfectly good finite limit while the plotted point is missing or placed elsewhere, creating a removable discontinuity rather than destroying the limit.
Discontinuities have distinct mechanisms
Removable, jump, infinite and oscillating discontinuities are not interchangeable labels. A removable defect has a shared finite limit and can be repaired at one point.
A jump has unequal one-sided limits. An infinite discontinuity has unbounded local behaviour, while an oscillating one never settles. Only the first admits a continuous extension by assigning the limiting value.
Intervals and composites carry conditions
Continuity on a closed interval uses a right-hand test at the left endpoint and a left-hand test at the right endpoint.
Sums, products and allowed powers of continuous functions remain continuous. Quotients require a nonzero denominator, roots require admissible inputs, and a composite requires the outer function to be continuous at the inner output.
State the valid interval rather than calling a formula continuous everywhere.
The Intermediate Value Theorem is an existence tool
A continuous function on a closed interval takes every output value between its endpoint outputs. For root questions, opposite endpoint signs guarantee at least one zero between them. The theorem does not locate the root and does not prove uniqueness.
A complete argument names continuity, computes both endpoint values, places the target between them and then invokes the theorem.
Separate repair from existence
Changing a single function value can repair only a removable discontinuity. It cannot join unequal side limits or control an unbounded or oscillating approach.
For interval theorems, establish continuity on the whole closed interval before using endpoint values to claim that an intermediate output occurs.
What this chapter covers
- 01
Value, limit and agreement at a point
- 02
Endpoint continuity on a closed interval
- 03
Four discontinuity mechanisms
- 04
Continuous extensions at removable holes
- 05
Closure rules for combinations and composites
- 06
Intermediate values and root existence
Worked example · free
Guarantee a root on a closed interval
- 2Establish continuity on the complete closed interval.
- 3Compute and compare the endpoint values with zero.
- 2Invoke the Intermediate Value Theorem with a modest conclusion.
Key terms
- Continuity Test
- The requirement that the point value exists, the limit exists and those values are equal.
- Removable Discontinuity
- A finite-limit defect repairable by assigning the limiting value at one point.
- Jump Discontinuity
- A point where finite left-hand and right-hand limits exist but are unequal.
- Continuous Extension
- A function that preserves the original rule and fills a missing point with its finite limit.
- Composite Continuity
- Continuity inherited when the inner function is continuous and the outer function is continuous at the inner output.
- Intermediate Value
- Any output between the endpoint outputs of a continuous function on a closed interval.
Continuity and the Intermediate Value Theorem FAQ
How do I test continuity at a piecewise join?
Find the left-hand limit, the right-hand limit and the defined value at the join. Continuity requires all three to be equal. When a parameter is unknown, this equality becomes the equation that determines it. Label the common value after solving.
Does a sign change always guarantee a root?
Only with continuity across the entire closed interval. A discontinuous function can jump from a negative value to a positive value without ever reaching zero. State why the function is continuous before computing endpoint signs and invoking the Intermediate Value Theorem.
Can the Intermediate Value Theorem prove a root is unique?
No. It guarantees at least one point where the target value occurs. A function may cross that value several times. Uniqueness needs additional evidence, such as strict monotonicity, that is separate from the theorem's existence conclusion.
Exam move
Build a discontinuity table with columns for point value, left limit, right limit, type and repair. Use it on graphs before attempting formulas. For composite and quotient questions, write the domain condition beside the continuity rule. Practise Intermediate Value Theorem proofs in four sentences: continuity, first endpoint value, second endpoint value and conclusion.
Circle any words such as unique or exact in your draft conclusion, because the theorem does not supply them. Rework piecewise parameter problems until value matching and slope matching remain separate tasks.
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