When Endpoint Values Do Not Tell the Whole Story
The Intermediate Value Theorem gives a powerful guarantee: if a function is continuous on a closed interval, then it attains every value between its endpoint values. The continuity condition is essential to that guarantee. A discontinuous function can have endpoint values on opposite sides of a target and still never take that target value.
A jump discontinuity makes this failure especially easy to see. The function can move through a range of values on one side of a point, then jump over a gap to a different range on the other side. The endpoint values may straddle a number in that gap, but no input produces it.
The Theorem and the Type of Discontinuity
The condition that \(f\) be continuous throughout the interval is not optional. Without it, the conclusion need not hold. We will use the following standard description of a jump at an interior point.
A jump creates a difference between the values approached from the left and from the right. That difference suggests a possible gap, but a jump at one point alone does not prove that a particular target is absent from the function’s entire range: some other input might produce it. To establish that a target is skipped, check every part of the function’s domain.
Worked Examples
Worked Example: A Two-Level Step Function
Define \(f:[-1,1]\to\mathbb{R}\) by
The endpoint values are \(f(-1)=-1\) and \(f(1)=2\). The target \(y=0\) lies strictly between them because \(-1<0<2\).
For every \(x<0\), the definition gives \(f(x)=-1\), not \(0\). For every \(x\ge 0\), it gives \(f(x)=2\), also not \(0\). Thus \(f(x)\ne 0\) for every \(x\in[-1,1]\). In fact, the whole range is \(\{-1,2\}\).
At the jump point, \(\lim_{x\to0^-}f(x)=-1\), while \(\lim_{x\to0^+}f(x)=2\), and \(f(0)=2\). The unequal one-sided limits show that \(f\) is discontinuous at \(0\). The endpoint check and the range check confirm the result: although the endpoints straddle \(0\), the function never attains \(0\).
Worked Example: A Jump Between Two Linear Pieces
Define \(g:[0,2]\to\mathbb{R}\) by
At the endpoints, \(g(0)=0\) and \(g(2)=3\), so \(0<1<3\). We will check whether the target \(1\) is attained.
On the first piece, \(0\le x<1\), so \(g(x)=x\) takes values in \([0,1)\). It does not equal \(1\), because \(x\) is strictly less than \(1\) on this piece. On the second piece, \(1\le x\le2\), so \(g(x)=x+1\) takes values in \([2,3]\). It cannot equal \(1\), because \(x+1\ge2\). Therefore the range is \([0,1)\cup[2,3]\), and \(1\) is missing.
At \(x=1\), the left-hand limit is \(1\), and the right-hand limit is \(2\). The defined value is \(g(1)=2\). The jump leaves a gap between the output values just below \(1\) and the outputs starting at \(2\); the target \(1\) is not included in either piece. Substituting the endpoints again gives \(g(0)=0\) and \(g(2)=3\), as claimed.
Worked Example: A Target Strictly Inside the Jump Gap
Define \(h:[-1,1]\to\mathbb{R}\) by
The endpoint values are \(h(-1)=2(-1)+1=-1\) and \(h(1)=2(1)+3=5\). Choose the target \(y=2\); it lies between the endpoints since \(-1<2<5\).
For \(-1\le x<0\), the first formula gives \(-1\le 2x+1<1\), so the outputs on this piece lie in \([-1,1)\). For \(0\le x\le1\), the second formula gives \(3\le2x+3\le5\), so the outputs lie in \([3,5]\). Neither interval contains \(2\). Thus \(h(x)\ne2\) for every \(x\in[-1,1]\), and its range is \([-1,1)\cup[3,5]\).
The one-sided limits at \(0\) are \(\lim_{x\to0^-}h(x)=1\) and \(\lim_{x\to0^+}h(x)=3\). The target \(2\) lies strictly between those limits, but the function has no values in the gap \([1,3)\). In particular, the endpoint values straddle \(2\) without the function attaining it. Direct substitution confirms the endpoint values used above.
A General Construction and Why It Works
The examples have a shared structure. We can prove directly that a simple family of jump-discontinuous functions always skips every target strictly between its two levels.
Let \(a<c<b\), and let \(L<R\). Define \(f:[a,b]\to\mathbb{R}\) by
First, because \(a<c\), the definition gives \(f(a)=L\). Since \(c\le b\), it also gives \(f(b)=R\). Thus the endpoint values are \(L<R\). Now take any target \(y\) satisfying \(L<y<R\). If \(a\le x<c\), then \(f(x)=L\), so \(f(x)\ne y\). If \(c\le x\le b\), then \(f(x)=R\), so again \(f(x)\ne y\). Those two pieces cover the entire domain, so no input can produce \(y\). At the jump point \(c\), the one-sided limits are \(L\) and \(R\), which are unequal. This proves both that the function has a jump and that it misses every target strictly between its endpoint values.
The crucial reasoning is the range check: every input belongs to one of the two pieces, and neither piece outputs the target. Merely noticing that the one-sided limits differ is not enough to establish that a particular value is missed; the full definition over the domain supplies that proof.
What the Examples Show About Continuity
These examples do not contradict the Intermediate Value Theorem. The theorem applies when the function is continuous on the whole closed interval. Each example has an interior jump discontinuity, so that hypothesis is false. The theorem therefore gives no conclusion about whether the target is attained.
It is useful to be precise about what “continuity is necessary” means here. Continuity is a necessary hypothesis for the theorem’s general guarantee: one cannot drop it and still claim that every function with endpoint values on opposite sides must attain every intermediate value. Continuity is not required for every individual function to attain intermediate values. A particular discontinuous function may still happen to take a particular target value. The examples show that, without continuity, the guarantee can fail.
A Common Mistake: Checking the Endpoints but Not the Hypotheses
Consider the first worked example. A mistaken argument might say, “\(f(-1)=-1<0<2=f(1)\), so the Intermediate Value Theorem gives an \(x\) with \(f(x)=0\).” The endpoint calculation is correct, but the conclusion does not follow: the function is discontinuous at \(0\). In fact, its definition shows that all its outputs are either \(-1\) or \(2\), so no such \(x\) exists.
The correction is to check both parts of an IVT argument: the endpoint values must straddle the target, and the function must be continuous on the full interval. If continuity fails, the endpoint comparison alone cannot establish that an intermediate value occurs. A discontinuous function may attain the target, miss it, or have more complicated behavior; the theorem simply does not settle the question.
A Reliable Check for a Skipped Value
When asked whether a discontinuous function misses a proposed intermediate value, proceed in a fixed order:
- Evaluate the function at both endpoints and verify that the target lies between those values.
- Locate the discontinuity and compute the one-sided limits if they exist.
- Use the piecewise definition to determine the outputs on every part of the domain.
- Check each piece against the target, including any point where the formula changes.
- State explicitly why no input in the domain gives the target.
The last two steps matter. A graph may suggest a gap, but a proof should account for every possible input, including the value assigned at the jump itself. For a piecewise function, that usually means finding the range of each piece and checking whether the target belongs to any of those ranges.
Check Your Understanding
For each question, explain the role of the jump and use the function’s definition to justify your conclusion.
- For \(f:[-1,1]\to\mathbb{R}\) defined by \(f(x)=-2\) when \(x<0\) and \(f(x)=3\) when \(x\ge0\), do the endpoint values straddle \(1\)? Is \(1\) attained?
- For \(g:[0,2]\to\mathbb{R}\) defined by \(g(x)=x\) when \(x<1\) and \(g(x)=x+1\) when \(x\ge1\), find the range of each piece. Which value between the endpoint values is missing?
- In the preceding function, what are the left-hand limit, right-hand limit, and function value at \(x=1\)?
- Why does a jump discontinuity at one point not, by itself, prove that a particular target is absent from the function’s entire range?
- A function has endpoint values \(0\) and \(4\), and has a jump at an interior point. Can the Intermediate Value Theorem be applied from those endpoint values alone? State the missing condition.
