Why Intermediate Values Matter
A continuous function cannot jump from one output value to another while skipping all the values in between. The Intermediate Value Theorem makes this idea precise: if a function is continuous on a closed interval, then its endpoint outputs determine a whole interval of values that the function must attain somewhere.
The theorem is an existence result. It tells us that an input with a specified output exists, but it does not usually identify that input, say that it is unique, or describe the function’s entire range. Those distinctions matter when deciding exactly what the theorem allows us to conclude.
Statement of the Intermediate Value Theorem
“Between” includes the endpoint values. If \(N=f(a)\), we can take \(c=a\); if \(N=f(b)\), we can take \(c=b\). The most informative case is when \(N\) lies strictly between the two endpoint outputs.
The hypothesis “continuous on \([a,b]\)” means that the function is continuous at every interior point and is continuous from within the interval at each endpoint. In particular, the function must be defined throughout the interval. Continuity at just the endpoints, or continuity only at most points, is not enough.
If the endpoint outputs are different, the theorem guarantees every output between them. In compact form, it guarantees that the function’s range contains the closed interval from the smaller endpoint output to the larger one:
This statement describes values the function must attain, not all values it attains. The function may take additional values outside that interval.
Applying the Theorem Carefully
A correct application checks the whole interval’s continuity, calculates both endpoint outputs, and then checks that the desired output lies between them. Only after those steps can we invoke the theorem.
Confirm that the interval is closed and that the function is continuous everywhere on it.
Compute \(f(a)\) and \(f(b)\) carefully.
Check whether \(N\) is between \(f(a)\) and \(f(b)\), inclusive.
Conclude that at least one \(c\in[a,b]\) satisfies \(f(c)=N\). Do not claim uniqueness unless you have a separate reason.
Worked Example: A Target Output Between the Endpoint Values
Let \(f(x)=x^2+1\) on \([0,2]\). Determine whether the theorem guarantees an input with output \(3\), and check the result.
First, \(f\) is a polynomial, so it is continuous on the entire closed interval \([0,2]\). The endpoint values are
The target \(3\) lies between them because \(1<3<5\). The Intermediate Value Theorem therefore guarantees at least one \(c\in[0,2]\) such that \(f(c)=3\).
Here we can also find and check such an input directly:
Only \(\sqrt{2}\) lies in \([0,2]\): it is positive, and \(2<4\) implies \(\sqrt{2}<2\). Substitution verifies the output:
The theorem’s conclusion is that at least one such input exists; the direct check confirms that \(\sqrt{2}\) is one. This example does not establish that the theorem generally finds an input or that such an input is unique.
Worked Example: Endpoint Values in Decreasing Order
Let \(g(x)=5-x^3\) on \([0,2]\). Does the theorem guarantee that \(g\) takes the value \(1\) on this interval?
The function is a polynomial, hence continuous on \([0,2]\). Evaluate both endpoints:
The endpoint outputs are in decreasing order, but that causes no problem: \(1\) is between \(-3\) and \(5\), since \(-3<1<5\). The theorem guarantees some \(c\in[0,2]\) with \(g(c)=1\).
For a check, solve the equation:
Since \(0<4<8\), taking cube roots gives \(0<\sqrt[3]{4}<2\), so this input is in the interval. Substitution confirms the output:
The conclusion is unchanged when the endpoint outputs are reversed: every output between them is still guaranteed.
Why the Theorem Works
The theorem formalizes the no-jump property of continuity. A short proof for one ordering of the endpoint values also shows why both continuity and the closed interval matter. Suppose \(f\) is continuous on \([a,b]\) and \(f(a)<N<f(b)\). Consider the inputs whose outputs are at most \(N\):
This set is nonempty because \(f(a)<N\), so \(a\in S\). It is bounded above by \(b\). Let \(c=\sup S\), the least upper bound of \(S\). Because \(f(a)<N\) and \(f\) is continuous at \(a\) from within the interval, some inputs immediately to the right of \(a\) also have outputs below \(N\). Thus \(c>a\). Likewise, \(f(b)>N\) and continuity from within at \(b\) imply that inputs sufficiently close to \(b\) have outputs above \(N\); hence \(c<b\).
We now show that \(f(c)=N\). If \(f(c)<N\), continuity at \(c\) implies that \(f(x)<N\) for all \(x\) sufficiently close to \(c\). In particular, there would be an \(x>c\) in \([a,b]\) with \(f(x)<N\). That would put \(x\) in \(S\), contradicting that \(c\) is an upper bound for \(S\).
If instead \(f(c)>N\), continuity at \(c\) implies that \(f(x)>N\) for all \(x\) sufficiently close to \(c\). But the definition of \(c=\sup S\) ensures that there are points of \(S\) arbitrarily close to \(c\) from the left: for every sufficiently small \(\varepsilon>0\), some \(s\in S\) satisfies \(c-\varepsilon<s\leq c\). Such points have \(f(s)\leq N\), a contradiction. Neither strict inequality is possible, so \(f(c)=N\).
This proves the theorem when \(f(a)<N<f(b)\). If \(f(b)<N<f(a)\), reverse the interval’s direction by defining \(h(t)=f(a+b-t)\) for \(t\in[a,b]\). This function is continuous, and its endpoint outputs are \(h(a)=f(b)<N<h(b)=f(a)\). The result just proved applies to \(h\); its guaranteed input corresponds to an input for \(f\) in \([a,b]\). When \(N\) equals an endpoint output, that endpoint itself supplies the required input. These cases establish the full conclusion.
What the Endpoint Information Does—and Does Not—Tell Us
Worked Example: Equal Endpoint Values Give a Limited Guarantee
Let \(q(x)=x^2\) on \([-2,2]\). This polynomial is continuous throughout the closed interval. Its endpoint values are
The only value between these equal endpoint outputs, inclusive, is \(4\). The Intermediate Value Theorem therefore guarantees that \(q\) takes the value \(4\), which is confirmed by \(q(-2)=4\) and \(q(2)=4\). It does not guarantee any other output from these endpoint values alone.
In fact, this particular function also takes every value from \(0\) to \(4\). For any \(y\in[0,4]\), the input \(x=\sqrt{y}\) lies in \([0,2]\), and
Also, \(x^2\geq0\) on \([-2,2]\), and \(x^2\leq4\) there, so it takes no values outside \([0,4]\). Thus its full range is \([0,4]\), even though equal endpoint values make the theorem’s endpoint-based guarantee only \(4\). The extra information comes from examining the function, not from the theorem’s endpoint conclusion.
More generally, the theorem does not say that a function is monotonic, that a given output occurs exactly once, or that the function’s entire range lies between its endpoint outputs. It guarantees at least one input for each value between the endpoint outputs. A function may attain the same value at several inputs, and it may take outputs outside that interval between the endpoints.
A Common Mistake: Ignoring Continuity
Endpoint calculations alone do not justify an application. For instance, a tempting but invalid argument is: “Let \(r(x)=1/x\) on \([-1,1]\). Then \(r(-1)=-1\) and \(r(1)=1\), so the Intermediate Value Theorem guarantees an input with output \(0\).”
The endpoint arithmetic is correct, but the theorem does not apply. The expression \(1/x\) is undefined at \(x=0\), so it does not define a function on all of \([-1,1]\), and it is not continuous on that interval. Moreover, \(1/x=0\) has no solution: for every \(x\) in its domain, \(1/x\neq0\). The correction is to check continuity on the entire closed interval before drawing any conclusion from the endpoint values.
Check Your Understanding
For each question, distinguish the theorem’s guaranteed conclusion from any additional information about the function.
- A function is continuous on \([1,4]\), with \(f(1)=-2\) and \(f(4)=6\). Which output values does the theorem guarantee it attains?
- If \(f(1)=6\) and \(f(4)=-2\), does the order of the endpoint outputs change which values are guaranteed?
- What does the theorem guarantee if \(f(a)=f(b)=3\)? Does it imply that \(f(x)=3\) for every \(x\in[a,b]\)?
- Why is it not enough to know that a function is defined and continuous at the two endpoints?
- If the theorem guarantees an input \(c\) with \(f(c)=N\), does it guarantee that there is only one such input?
