Math Core

If you drive from sea level up to a mountain town at $2{,}000$ meters, at some moment you must have been at exactly $1{,}000$ meters. You cannot skip an altitude, because altitude changes continuously. The intermediate value theorem turns this common-sense idea into a precise statement, and it is the first of several existence theorems in AP Calculus that rely on continuity.

The theorem

The intermediate value theorem (IVT)

If $f$ is continuous on the closed interval $[a, b]$ and $k$ is any number between $f(a)$ and $f(b)$ , then there is at least one number $c$ in $[a, b]$ such that

f(c) = k.

(If $k$ is strictly between $f(a)$ and $f(b)$ , then $c$ can be taken strictly between $a$ and $b$ .)

Graphically: draw any horizontal line $y = k$ between the heights $f(a)$ and $f(b)$ . A continuous graph that starts on one side of that line and ends on the other must cross it at least once.

A continuous f on [a, b] with f(a) below k and f(b) above k. The graph crosses the dashed line y = k, here three times.Open in grapher →

Three features of the theorem are worth stressing.

Continuity is essential. If $f$ has a break anywhere in $[a, b]$ , the conclusion can fail.
It guarantees existence, not location. The IVT says a value $c$ exists; it does not tell you what $c$ is.
"At least one." There may be several such $c$ values, as in the graph above.

Why continuity matters

Consider $f(x) = \dfrac{1}{x}$ on $[-1, 1]$ . We have $f(-1) = -1$ and $f(1) = 1$ , and $0$ is between them. But $\dfrac{1}{x}$ is never $0$ . The IVT doesn't apply because $f$ is not continuous on $[-1, 1]$ : it is undefined at $x = 0$ , where the graph jumps from $-\infty$ to $\infty$ .

Or consider a piecewise function that is $-1$ for $x < 0$ and $1$ for $x \ge 0$ , on $[-2, 2]$ . It never equals $\dfrac{1}{2}$ , again because of its jump.

Using the IVT to locate roots

The most common application: if a continuous function changes sign on an interval, it has a zero there.

Worked example: Showing that an equation has a solution

Show that $x^3 + x - 5 = 0$ has a solution between $x = 1$ and $x = 2$ .

Solution. Let $f(x) = x^3 + x - 5$ . As a polynomial, $f$ is continuous on $[1, 2]$ . Also

f(1) = 1 + 1 - 5 = -3 \quad \text{and} \quad f(2) = 8 + 2 - 5 = 5.

Since $f(1) = -3 < 0 < 5 = f(2)$ and $f$ is continuous on $[1, 2]$ , the intermediate value theorem guarantees a number $c$ in $(1, 2)$ with $f(c) = 0$ .

This is how an AP free-response justification should look: name the function, state that it is continuous on the closed interval (with a reason), show the values at the endpoints, state that $k$ is between them, and conclude by the IVT.

You can repeat the process to narrow things down. Since $f(1.5) = 3.375 + 1.5 - 5 = -0.125 < 0$ and $f(2) > 0$ , the root is actually in $(1.5, 2)$ . Halving the interval over and over like this (the bisection method) pins the root down to any accuracy you want.

Worked example: Equations with a function on each side

Show that $\cos x = x$ has a solution in $\left[0, \dfrac{\pi}{2}\right]$ .

Solution. Move everything to one side: let $g(x) = \cos x - x$ , which is continuous everywhere as a difference of continuous functions. Then

g(0) = 1 - 0 = 1 > 0 \quad \text{and} \quad g\left(\frac{\pi}{2}\right) = 0 - \frac{\pi}{2} < 0.

By the IVT, $g(c) = 0$ for some $c$ in $\left(0, \dfrac{\pi}{2}\right)$ , which means $\cos c = c$ .

The IVT with tables

AP questions often give a table of values of a continuous function and ask what the IVT guarantees.

Worked example: Counting guaranteed solutions from a table

The function $f$ is continuous on $[0, 8]$ , and selected values are shown.

$x$	$0$	$2$	$5$	$8$
$f(x)$	$1$	$6$	$-2$	$4$

What is the fewest number of solutions the equation $f(x) = 3$ must have on $[0, 8]$ ?

Solution. Look at each consecutive pair of values and ask whether $3$ lies between them.

On $[0, 2]$ : $1 < 3 < 6$ . Yes.
On $[2, 5]$ : $-2 < 3 < 6$ . Yes.
On $[5, 8]$ : $-2 < 3 < 4$ . Yes.

The intervals $(0, 2)$ , $(2, 5)$ and $(5, 8)$ don't overlap, so there are at least $3$ different solutions.

Common mistake

The IVT works in only one direction. If $k$ is between $f(a)$ and $f(b)$ , a solution is guaranteed. If $k$ is not between them, the IVT says nothing: $f$ might still equal $k$ somewhere in the interval (for example, $f(x) = x^2$ on $[-1, 1]$ has $f(-1) = f(1) = 1$ but also reaches $0$ ). And always check continuity first; without it, no conclusion is possible.

Tip

To use the IVT for an equation like $\cos x = x$ or $e^x = 3 - x$ , rewrite it as (something) $= 0$ and apply the theorem to the difference. Sign changes are much easier to spot than crossings of two curves.

Practice

Practice 1

Let $f(x) = x^3 - 4x + 1$ . On which interval does the intermediate value theorem guarantee that $f$ has a zero?

Practice 2

The function $f$ is continuous on $[0, 4]$ , with values shown in the table.

$x$	$0$	$1$	$2$	$3$	$4$
$f(x)$	$3$	$-1$	$2$	$4$	$-2$

What is the least number of zeros $f$ must have on $[0, 4]$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

The function $g$ is continuous on $[1, 5]$ with $g(1) = 4$ , $g(3) = -2$ , and $g(5) = 6$ . What is the least number of solutions of $g(x) = 1$ on $[1, 5]$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 4

Let $f(x) = \dfrac{1}{x - 2}$ . Then $f(1) = -1$ and $f(3) = 1$ , but there is no $c$ in $[1, 3]$ with $f(c) = 0$ . Why doesn't this contradict the intermediate value theorem?

Practice 5

Let $f(x) = x^2 + x$ . The IVT guarantees a value $c$ in $[0, 3]$ with $f(c) = 6$ . Find $c$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 6

Which is a correct justification that $e^x = 3 - x$ has a solution in $[0, 1]$ ?

Practice 7

The function $g$ is continuous on $[-2, 4]$ with $g(-2) = 5$ and $g(4) = -3$ . Which statement must be true?