Intermediate Value Proof Workshop

From an Existence Question to a Bracket

The Intermediate Value Theorem is an existence tool: continuity and endpoint values guarantee that a function takes every value between them. In a proof, the main work is often not invoking the theorem but choosing the right function and verifying the hypotheses. The previous tutorial described the structure of connected subsets of \(\mathbb{R}\); here we use the interval form of the Intermediate Value Theorem to build practical arguments.

Suppose the goal is to show that \(f(x)=y\) for some \(x\) in an interval. The direct strategy is to define a new function \(h(x)=f(x)-y\). Then \(f(x)=y\) is equivalent to \(h(x)=0\). If \(h\) is continuous and its values at two points have opposite signs, the Intermediate Value Theorem gives the desired zero. This reformulation is useful because many existence questions become sign checks at two carefully chosen points.

The bracket endpoints need not be close to the unknown solution. They need only belong to the function's domain, have the required signs, and enclose an interval on which the function is continuous. If an endpoint already gives zero, the existence conclusion is immediate; a strict sign-changing bracket is particularly useful when the goal is to locate a zero strictly between the endpoints.

Three Ways to Set Up the Function

Before applying the theorem, identify exactly what equation the desired point must satisfy. For a target equation \(f(x)=y\), use \(h(x)=f(x)-y\). For an equation \(f(x)=g(x)\), use \(h(x)=f(x)-g(x)\). For a claimed equality between an unknown quantity and an expression depending on that quantity, move all terms to one side. In each case, the target becomes \(h(x)=0\).

A Comparison Principle for Two Functions

The same setup handles equations in which two functions must agree. Instead of trying to solve \(f(x)=g(x)\) directly, examine their difference. The following result packages this method as a useful existence principle.

Proof. Define \(h:J\to\mathbb{R}\) by \(h(x)=f(x)-g(x)\). Since \(f\) and \(g\) are continuous, their difference \(h\) is continuous on \(J\), and therefore on \([a,b]\). In the first endpoint configuration, \(f(a)<g(a)\) gives \(h(a)<0\), while \(f(b)>g(b)\) gives \(h(b)>0\). The Intermediate Value Theorem supplies \(c\in(a,b)\) with \(h(c)=0\). Consequently \(f(c)-g(c)=0\), so \(f(c)=g(c)\). In the second configuration, the endpoint signs are reversed, and the same theorem again gives a zero in \((a,b)\). This proves the result. \(\square\)

The strict inequalities ensure that the meeting point lies in the open interval. If the functions are already equal at an endpoint, there is an equality on \([a,b]\), but that equality need not occur in \((a,b)\).

Application: Every Odd-Degree Polynomial Has a Real Root

A useful global application combines endpoint bracketing with the leading term of a polynomial. The leading term controls the signs far enough to the left and right. Once those signs are opposite, continuity supplies a root between the chosen endpoints.

Proof. Let

where \(n\) is odd and \(a_n\ne0\). Write \(r(x)=a_{n-1}x^{n-1}+\cdots+a_1x+a_0\), so \(p(x)=a_nx^n+r(x)\). Set \(C=\sum_{k=0}^{n-1}|a_k|\). For any \(R\geq1\), each power \(R^k\), with \(0\leq k\leq n-1\), is at most \(R^{n-1}\). Therefore

Choose \(R\geq1\) so large that \(R>C/|a_n|\). Then \(CR^{n-1}<|a_n|R^n\). At \(R\), the remainder has magnitude smaller than the leading term, so \(p(R)\) has the same sign as \(a_nR^n\), hence the same sign as \(a_n\). At \(-R\), because \(n\) is odd, \((-R)^n=-R^n\). The leading term \(a_n(-R)^n\) therefore has sign opposite to \(a_n\); the smaller remainder cannot change that sign. Thus \(p(-R)\) and \(p(R)\) have opposite signs.

Polynomials are continuous on \([-R,R]\). Applying the Intermediate Value Theorem to \(p\) gives \(c\in(-R,R)\) with \(p(c)=0\). Hence \(p\) has a real root. \(\square\)

What the Argument Does—and Does Not—Show

A sign-changing bracket proves that at least one zero lies between its endpoints. It does not, by itself, prove that there is exactly one zero. A continuous function may cross zero more than once inside a bracket. To establish uniqueness, an additional argument is needed, such as proving strict monotonicity or using another structural property of the equation.

There are also two hypothesis checks that should not be skipped. First, the function used in the argument must actually be continuous on the whole interval between the endpoints, not merely at the endpoints. Second, the endpoint values must be computed for the function that encodes the target equation. For instance, when the goal is \(f(x)=y\), checking the signs of \(f(a)\) and \(f(b)\) alone is not enough; the relevant signs are those of \(f(a)-y\) and \(f(b)-y\).

When selecting endpoints, exact values are ideal, but estimates can suffice if they rigorously establish opposite signs. The essential proof pattern is to identify the zero-equivalent function, justify continuity, and verify the endpoint inequalities. The Intermediate Value Theorem then supplies the point whose existence was required.