# Intermediate Value Theorem

The **intermediate value theorem** states that if a continuous function attains two values, it must also attain all values in-between these two values. Intuitively, a continuous function is a function whose graph can be drawn without lifting pencil from paper. For instance, if \(f(x)\) is a continuous function that connects the points \([0,0]\) and \([5,20]\), then there must be some \(x_a\) where \(y=1\) also some \(x_b\) where \(y=19\), another \(x_c\) where \(y= \frac{19}{5}\), etc. For this function there is an \(x\) for any \(y\) value between \(0\) and \(20\).

Formally, for two reals \(a\) and \(b\) with \(a < b\), a function \(f\) is continuous on a closed interval [a, b] such that \(f(a)<f(b)\). Then for every \(y_0\) such that \(f(a)<y_0<f(b)\) there exists a number \(x_0\ ∈[a, b]\) with \(f(x_0)=y_0\).

More precisely, if \(f: [a,b] \to \mathbb{R}\) is continuous, then for each \(c\) such that \[\min(f(a), f(b)) \le c \le \max(f(a), f(b)), \] there exists \(x\in [a,b]\) with \(f(x) = c\).

Suppose then that two points \((a,f(a))\) and \((b,f(b))\) are on the graph of some continuous function \(f: [a,b] \to \mathbb{R}\). These points demarcate a segment of the graph which connects the two points, as in the picture above. Since this segment has no holes or breaks in it (by continuity of the function), it must pass through each "level curve" \(y=c,\) where \(\min(f(a), f(b)) \le c \le \max(f(a), f(b)) \).

## Proof

An intuitive proof of the theorem goes as follows:

Suppose the segment does not pass through some line \(y=c\). Assume without loss of generality that \(f(a) < c < f(b)\). Then, the graph of \(f\) must stay below the line \(y=c\) at all \(x\in [a,b]\), so \(f\) cannot attain the value \(f(b)\). However, we already know it does, so this is absurd. This argument can be made rigorous using the notions of *connected sets* and *open sets* from general topology.

## Common Misinterpretations

Because the statement of the theorem has multiple requirements, one may overlook one of the premises as in the following example:

For the function \(f(x) = x^2 - \dfrac{9}{x} + 1\), over which of the following intervals does the intermediate value theorem guarantee a root:

\[\begin{array} &\left[-3, -1\right], &\left[-1, 1\right], &\left[1, 3\right]? \end{array}\]

In order for the intermediate value theorem to guarantee a root on a specified interval \(\left[a, b\right]\), not only must the function \(f\) be continuous on the interval, but 0 must be contained between \(f(a)\) and \(f(b).\) Let's check the values of \(f(-3), f(-1), f(1),\) and \(f(3):\)

\[\begin{array} &f(-3) = 13, &f(-1) = 11, &f(1) = -7, &f(3) = 7. \end{array} \]

For the first interval \(\left[-3, -1\right],\) the values returned by \(f\) are both positive which do

notsandwich 0, meaning the intermediate value theorem does not guarantee a root.For the second interval \(\left[-1, 1\right],\) the values returned by \(f\) are on either side of 0, which seems to suggest that \(f\) has a root on the interval \(\left[-1, 1\right].\) However, it's important to note that \(f(x)\) has a discontinuity at \(x = 0,\) meaning the intermediate value theorem does not hold. Indeed, \(f(x)\) does not have an \(x\)-intercept on the interval \(\left[-1, 1\right].\)

For the last interval \(\left[1, 3\right],\) the values returned by \(f\) are on either side of 0, which implies that \(f\) has a root on the interval \(\left[1, 3\right].\) This is confirmed by the intermediate value theorem because \(f\) is continuous on \(\left[1, 3\right].\) \(_\square\)

## Examples and Applications

One simple and important use of the intermediate value theorem (hereafter referred to as IVT) is to prove that certain equations have solutions. Consider the following example:

Does the equation \(\cos(x) = x\) have solution(s) \(x\in \mathbb{R}\)? If so, how many solutions does it have?

We study the function \(f(x) = x-\cos x\). Note that \(f(0) = -1 < 0\) and \(f(\pi) = \pi + 1 > 0\). Thus, by the IVT, there must be some \(y\in [0,\pi]\) such that \(f(y) = 0 \), i.e. \(y = \cos y\).

One root of the equation has been identified. Is this the only root? Note that \(f'(x) = 1-\sin x\) is everywhere non-negative, so \(f\) is increasing monotonically. Hence, \(f\) can only have one root. \(_\square\)

Suppose that \(f\) is continuous on \([0, 1]\) and \(f(0) = f(1)\). Let \(n\) be any positive integer, then prove that there is some number \(x\) such that \[f(x) = f\left(x +\dfrac{1}{n}\right).\]

Define \(g(x) = f(x) − f\left(x +\dfrac{1}{n}\right)\).

Consider the set of numbers \(S =\left\{f(0), f\left(\dfrac{1}{n}\right), f\left(\dfrac{2}{n}\right), . . . , f(1)\right\}\).

Let \(k\) be such that \(f\left(\dfrac{k}{n}\right)\) is the largest number in \(S\). Suppose that \(k \ne 0\) and \(k \ne n\).

Then \(g\left(\dfrac{k}{n}\right) = f\left(\dfrac{k}{n}\right) − f\left(\dfrac{k+1}{n}\right) \geq 0\), and \(g\left(\dfrac{k-1}{n}\right) = f\left(\dfrac{k-1}{n}\right) − f\left(\dfrac{k}{n}\right)\leq 0\).

By the Intermediate value theorem, there is \(c\in \left[\dfrac{k-1}{n},\dfrac{k}{n}\right]\) with \(g(c) = 0\), so that \(f(c) − f\left(c + \dfrac{1}{n}\right) = 0\), or \(f(c) − f\left(c + \dfrac{1}{n}\right)\) as desired.

Finally, if the largest number in \(S\) is \(f(0) = f(1)\), then the same argument works with \(k\) chosen such that \(f\left(\dfrac{k}{n}\right)\) is the minimum number in \(S\). \(_\square\)

Note that if \(f(0)\) is both the largest and smallest number in \(S\), then they are all the same and \(f(0) = f\left(\dfrac{1}{n}\right)\).

Of the four statements below, which are true?

**I.** The equation \(x^4 -3x+ 1 = 0\) has a unique real solution.

**II.** The equation \(\sin x = x\) has a unique real solution.

**III.** The equation \(3x^5 - 20x^3 + 60x +16=0\) has a unique real solution.

**IV.** The equation \(\tan x = x\) has a unique real solution.

A function \(f: [a,b] \to \mathbb{R}\) (where \(a\), \(b\) might equal \(\pm \infty\)) is said to have the *intermediate value property* if for every \[x \in [\min(f(a), f(b)), \max(f(a), f(b))],\] there exists some \(c\in [a,b]\) such that \(f(c) = x\).

The intermediate value theorem states that if \(f\) is continuous, then \(f\) has the intermediate value property. Is the converse of this theorem true? That is, if a function has the intermediate value property, must it be continuous on its domain?

Since it can detect zeroes of functions, the IVT is an important tool for the analysis of continuous functions. However, through some clever contortions, IVT can give even more impressive results. For instance, one can prove the **Borsuk-Ulam theorem** in dimension 1. This theorem states that for any continuous real-valued function \(f\) on a circle, there is some point \(p\) on the circle such that \(f\) takes the same value at \(p\) and at the point on the circle directly opposite to \(p\) (the *antipode* of \(p\)).

This implies that on any great circle of the globe, any continuously varying information will take on the same value at some two antipodal points. For instance, there must exist two antipodal points on the equator at which the air temperature is the same.

Let \(S^1\) denote a circle, and suppose that \(f: S^1 \to \mathbb{R}\) is a continuous function. Then, there exists \(\mathbf{x}\in S^1\) such that \(f(\mathbf{x}) = f(-\mathbf{x})\). \(_\square\)

Note: Here, by the circle \(S^1\), we mean the set of vectors in \(\mathbb{R}^2\) of length precisely 1. In other words, \(S^1\) is the set of points \((x,y) \in \mathbb{R}^2\) such that \(x^2 + y^2 = 1\).

There is a function \(p: [0,2\pi) \to S^1\) given by \(p(\theta) = (\cos \theta, \sin \theta)\). Composing this with \(f\) gives \(g:= f\circ p : [0,2\pi) \to \mathbb{R}\). Define a function \(h:[0,2\pi) \to \mathbb{R}\) by \[h(\theta) := g(\theta + \pi) - g(\theta),\] where we take \(\theta + \pi \pmod{2\pi}\) if \(\theta >\pi\).

Note that \[h(0) = g(\pi) - g(0) = -(g(0) - g(\pi)) = -h(\pi).\] In particular, \(h(0)\) and \(h(\pi)\) have opposite signs. Thus, by the IVT, there is some \(t \in (0,\pi)\) such that \(h(t) = 0\). This means \(g(t + \pi) = g(t)\).

Let \(\mathbf{x} = (\cos t, \sin t)\). Since \(p(t+\pi) = -\mathbf{x}\), we conclude \[f(-\mathbf{x}) = g(t+\pi) = g(t) = f(\mathbf{x})\] as desired. \(_\square\)

Any *die* is modeled by some polyhedron. If the polyhedron is *completely symmetric* in the sense that any face can be taken to any other face via a rigid motion, then the die will be *fair*; when the die is rolled, the probability of landing on any face will equal the probability of landing on any other face.

Do there exist fair dice that are not completely symmetric?

**Hint**: Start with a prism \(P\) whose cross-sections are regular \(n\)-gons. Now consider the *dual* polyhedron \(P^{\star}\), the polyhedron whose vertices are the centers of the faces of the original prism. This \(P^{\star}\) looks like two pyramids with regular \(n\)-gon cross-sections that have been glued together at their bases. Now, can you modify \(P^{\star}\) to obtain a fair die that isn't completely symmetric?

Is there a solution to \(x^5 - 2 x^3 - 2 = 0,\) where \(x\in [0,2]\)?

At \(x=0\), we have \(0^5 - 2 \times 0^3 - 2 = -2.\)

At \(x=2\), we have \(2^5 - 2 \times 2^3 - 2 = 14.\)

Now we know that at \(x=0\) the curve is below zero, and that at \(x=2\) the curve is above zero. Being a polynomial, the curve will be continuous, so somewhere in between, the curve must cross through \(y=0.\)

Therefore, there is a solution to \(x^5 - 2 x^3 - 2 = 0\) in the interval \( [0, 2]\). \(_\square\)

**Cite as:**Intermediate Value Theorem.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/intermediate-value-theorem/