Prove a Continuous Function Attains All Points on a to B

The statement of the theorem has multiple requirements, all of which are necessary for the conclusion to hold. Here is an illustrative example:

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

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

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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 $f(3):$

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

For the first interval $\left[-3, -1\right],$ the values returned by $f$ are both positive which do not sandwich 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 $\left[1, 3\right].$ $_\square$

Graph of f(x)

As in the above example, 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?

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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. $_\square$

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

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At $x=0$ , we have $0^5 - 2 \times 0^3 - 2 = -2.$
At $x=2$ , we have $2^5 - 2 \times 2^3 - 2 = 14.$

So the IVT implies that there is a solution to $x^5 - 2 x^3 - 2 = 0$ in the interval $[0, 2]$ . $_\square$

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

$f(x) = f\left(x +\dfrac{1}{n}\right).$

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Define $g(x) = f(x) - f\left(x +\frac{1}{n}\right)$ .

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

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

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

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

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

Suppose that $f$ is continuous on $[0, 1]$ and $f(0) = f(1)$ . Let $\varepsilon$ be the hyperreal unit, then prove that there is some number $x$ such that

$f(x) = f\left(x +\varepsilon \right).$

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First, assume $f(x)$ is not constant on $[0,1]$ . The result holds trivially if it is.

Then, let $g(x)=f(x)-f(x+\varepsilon)$

Since $f$ is not constant, there exists a $c\in[0,1]$ such that $f(c)$ is a maximum (or a minimum, but assume for now that it is a max; a min is handled similarly).

Then, $g(c) = f(c) - f(c+\varepsilon) \geq 0$ and $g(c-\varepsilon) = f(c-\varepsilon) - f(c) \leq 0$

Now, by the Intermediate Value Theorem, if $\varepsilon >0$ , there exists a $y\in[c-\varepsilon,c]$ such that $g(y)=0$ .

$($ If $\varepsilon < 0, y\in[c,c-\varepsilon]$ , instead. $)$

Thus, $f(y) = f(y+\varepsilon)$ , as wanted.

If $\varepsilon = 0$ , $f(x) = f(x+\varepsilon)\ \forall x$ . $_\square$

I, II, and III I and III II and III II, III, and IV

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.

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 $f(\mathbf{x}) = f(-\mathbf{x})$ .

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

$h(0) = g(\pi) - g(0) = -\big(g(0) - g(\pi)\big) = -h(\pi).$

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?

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Source: https://brilliant.org/wiki/intermediate-value-theorem/

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