Intermediate Value Theorem

Statement

1.Let $f : [a,b] \to \mathbb{R}$ be continuous on the closed interval $[a,b],$ and let $y$ lie between the endpoint values, so that $f(a) \le y \le f(b)$ or $f(b) \le y \le f(a)$. Then at least one point $c \in [a,b]$ satisfies:

$$f(c) = y$$

2.Every horizontal line whose height is between $f(a)$ and $f(b)$ meets the graph of $f$ above the interval $[a,b]$.

3.The theorem is an existence result. It neither locates $c$ nor guarantees that $c$ is unique. An intermediate value can be attained once or several times. The function $f(x) = x^3 - 4x$ on $[-3,3]$ has $f(-3) = -15$ and $f(3) = 15,$ while the value $0$ is attained at the three points $x = -2,$ $x = 0$ and $x = 2.$

The zero theorem of Bolzano

4.For $y = 0,$ the intermediate value theorem has the following form. Suppose that $f$ is continuous on $[a,b]$ and that its endpoint values have opposite signs, so that:

$$f(a)f(b) < 0$$

5.Then $f$ has a zero at some point $c$ of the open interval $(a,b)$. This result is known as the zero theorem or Bolzano's theorem. Bolzano published an analytic proof in 1817 without appealing to a graph.

6.The hypotheses impose no bound on the number of zeros, and a continuous function can have infinitely many zeros.

Proof by successive bisection

7.Assume $f(a) < 0 < f(b)$. If the signs are reversed, replace $f$ with $-f,$ which has the same zeros.

8.Put $a_0 = a$ and $b_0 = b$. Suppose that the interval $[a_n,b_n]$ has been constructed with $f(a_n) < 0 < f(b_n)$. Its midpoint is:

$$m_n = \frac{a_n + b_n}{2}$$

9.If $f(m_n) = 0,$ then $m_n$ is a zero and the construction stops. If $f(m_n) > 0,$ keep the left half by setting $a_{n+1} = a_n$ and $b_{n+1} = m_n$. If $f(m_n) < 0,$ keep the right half by setting $a_{n+1} = m_n$ and $b_{n+1} = b_n$. In either case the new endpoints have opposite signs:

$$f(a_{n+1}) < 0 < f(b_{n+1})$$

10.Each new interval has half the length of the preceding interval. After $n$ steps its length is:

$$b_n - a_n = \frac{b-a}{2^n}$$

11.If the construction never stops, $(a_n)$ is increasing and $(b_n)$ is decreasing, and both sequences are contained in $[a,b]$. Every bounded monotone sequence converges, so both sequences have a limit. Their limits are equal because $b_n - a_n \to 0$. Denote their common limit by $c$. Since $[a,b]$ is closed, $c \in [a,b]$.

12.

The convergence step uses the completeness of the real numbers. Over $\mathbb{Q},$ the intervals can converge in $\mathbb{R}$ to an irrational number that is absent from the domain. For example, the restriction of $f(x) = x^2 - 2$ to $[1,2] \cap \mathbb{Q}$ is continuous and changes sign, but it has no rational zero.

13.Since $f$ is continuous at $c,$ we have $f(a_n) \to f(c)$ and $f(b_n) \to f(c)$. Every term of $(f(a_n))$ is negative, so $f(c) \le 0$. Every term of $(f(b_n))$ is positive, so $f(c) \ge 0$. Hence:

$$f(c) = 0$$

14.

The strict inequalities become weak inequalities in the limit. From $f(a_n) < 0$ we may conclude only that $f(c) \le 0,$ so neither sequence settles the matter alone.

15.Another proof uses the set $S = \{\ x \in [a,b] \mid f(x) < 0\ \}$. This set is nonempty because $a \in S$ and is bounded above by $b,$ so it has a supremum $c$. Continuity near $b$ shows that $c < b$. If $f(c) < 0,$ continuity gives a point of $S$ greater than $c,$ contradicting the definition of $c$. If $f(c) > 0,$ continuity gives an interval to the left of $c$ that contains no point of $S$. A number smaller than $c$ would then be an upper bound for $S,$ again a contradiction. Hence $f(c) = 0$. In the bisection proof, $c$ is also the limit of nested intervals that can be computed.

From zeros to arbitrary values

16.The zero theorem implies the general statement. If $y = f(a),$ take $c = a$; if $y = f(b),$ take $c = b$. It remains to consider strict inequalities. First suppose $f(a) < y < f(b)$ and define:

$$g(x) = f(x) - y$$

17.The constant function with value $y$ is continuous, so $g$ is continuous on $[a,b]$. Its endpoint values satisfy $g(a) = f(a) - y < 0$ and $g(b) = f(b) - y > 0$. Bolzano's theorem gives a point $c \in (a,b)$ with $g(c) = 0,$ or equivalently $f(c) = y$. The same argument applies when $f(b) < y < f(a),$ with the endpoint signs reversed.

18.Conversely, the zero theorem is the case $y = 0$ of the intermediate value theorem. The two statements are therefore equivalent.

Where the hypotheses enter

19.The function must be continuous on its whole domain, and the domain must be an interval. If either condition is removed, the conclusion can fail.

20.A counterexample on $[-1,1]$ is a variant of the Heaviside step function:

$$ f(x) = \begin{cases} -1 & x \le 0 \\[6pt] 1 & x > 0 \end{cases} $$

21.The endpoint values are $f(-1) = -1$ and $f(1) = 1,$ yet the function takes no value strictly between them. It differs from the sign function only at the origin, where it has a jump discontinuity. Its one-sided limits at the origin exist and are different.

22.Continuity alone is not enough when the domain is disconnected. The function $f(x) = 1/x$ is continuous at every point of $[-1,0) \cup (0,1]$ and satisfies $f(-1) = -1$ and $f(1) = 1,$ but it never vanishes. Its domain is the union of two disjoint intervals. On each interval the function has a constant sign, so applying the theorem separately to the two intervals gives no contradiction.

23.The converse is false. A function can have the intermediate value property on every interval and still be discontinuous. For example, $g(x) = \sin(1/x)$ for $x \neq 0,$ with $g(0) = 0,$ has this property but is discontinuous at the origin. Darboux's theorem states that every derivative has the intermediate value property, even when the derivative is discontinuous.

The image of an interval

24.If $f$ is continuous on an interval $I,$ then the image $f(I)$ is again an interval.

25.Let $u,v \in f(I)$ with $u < v$. Choose $p,q \in I$ such that $u = f(p)$ and $v = f(q),$ and suppose $u < y < v$. The closed interval with endpoints $p$ and $q$ is contained in $I$ because $I$ contains every point between any two of its points. The restriction of $f$ to this closed interval satisfies the hypotheses of the theorem. Hence $f(c) = y$ for some point $c$ between $p$ and $q,$ so $y \in f(I)$. Thus $f(I)$ contains every number between any two of its elements and is an interval.

26.If $f$ is continuous on the closed and bounded interval $[a,b],$ then Weierstrass' theorem shows that $f$ has a minimum $m$ and a maximum $M$. Thus $f([a,b])$ is an interval contained in $[m,M]$ and containing both $m$ and $M$. Therefore:

$$f([a,b]) = [m,M]$$

27.

The image need not be an interval of the same type as the domain. The map $f(x) = x^2$ sends the open interval $(-1,1)$ to $[0,1),$ which is neither open nor closed. If the domain is compact, its continuous image is compact and hence is a closed and bounded interval.

Example 1

28.Consider the polynomial function $f : [1,2] \to \mathbb{R}$ defined by:

$$f(x) = x^3 - x - 1$$

29.A polynomial function is continuous on the whole real line, so $f$ is continuous on $[1,2]$. Its endpoint values are:

$$ \begin{align} f(1) &= 1 - 1 - 1 = -1 \\[6pt] f(2) &= 8 - 2 - 1 = 5 \end{align} $$

30.The endpoint values have opposite signs, so the equation $x^3 - x - 1 = 0$ has at least one solution in $(1,2)$. Bisection narrows its location by retaining the half whose endpoints have opposite signs:

$$ \begin{align} f(1.5) &= 0.875 > 0 \quad \rightarrow \quad c \in (1,1.5) \\[6pt] f(1.25) &= -0.296875 < 0 \quad \rightarrow \quad c \in (1.25,1.5) \\[6pt] f(1.375) &= 0.224609375 > 0 \quad \rightarrow \quad c \in (1.25,1.375) \end{align} $$

31.After three steps the root belongs to an interval of length $0.125$. After $n$ steps the interval that contains the root has length $1/2^n,$ so its width can be prescribed in advance. Since $f'(x) = 3x^2 - 1 > 0$ on $[1,2],$ the function is strictly increasing there and the root is unique. This root is the plastic number, whose value to five decimal places is $1.32472$.

Real roots of polynomials of odd degree

32.A polynomial with real coefficients and odd degree has at least one real root.

33.Let $p(x) = a_nx^n + \dots + a_0,$ where $n$ is odd and $a_n \neq 0$. If $a_n < 0,$ replace $p$ with $-p,$ which has the same roots. We may therefore assume $a_n > 0$. The following limit compares $p$ with its leading term:

$$\lim_{|x| \to \infty} \frac{p(x)}{a_nx^n} = 1$$

34.The limit shows that the leading term determines the sign of $p(x)$ when $|x|$ is large. Since $n$ is odd, $p(x) \to +\infty$ as $x \to +\infty$ and $p(x) \to -\infty$ as $x \to -\infty$. Hence some points $u < v$ satisfy $p(u) < 0 < p(v)$. Bolzano's theorem applied to $p$ on $[u,v]$ gives a real root.

35.If the degree is even and the leading coefficient is positive, both limits are $+\infty,$ so no sign change follows. For example, $p(x) = x^2 + 1$ has no real root. The number and location of roots are studied further in polynomial equations.

Fixed points of a continuous self-map

36.Let $f : [a,b] \to [a,b]$ be continuous. Then $f$ has a fixed point, meaning a point $x_0$ with $f(x_0) = x_0$.

37.Since $f$ has values in $[a,b],$ define the continuous function $g(x) = f(x) - x$ on $[a,b]$. The inclusion $f(a) \in [a,b]$ gives $f(a) \ge a$ and hence $g(a) \ge 0$. Similarly, $f(b) \in [a,b]$ gives $f(b) \le b$ and hence $g(b) \le 0$. If either endpoint value of $g$ is zero, the corresponding endpoint is a fixed point. Otherwise $g(a) > 0 > g(b),$ and the zero theorem gives a point $x_0 \in (a,b)$ with $g(x_0) = 0$. Therefore $f(x_0) = x_0$.

38.The graph of $f$ lies in the square $[a,b]^2,$ joins its left and right edges, and intersects the diagonal. This result is the one-dimensional case of Brouwer's fixed point theorem for continuous maps from a closed ball of $\mathbb{R}^n$ into itself.

Where this theorem is used