The Intermediate Value Theorem

If the temperature was $-4$°C at 6 a.m. and $8$°C at 2 p.m., then at some moment in between it was exactly $0$°C. Temperature can’t skip a value, because it changes continuously. The Intermediate Value Theorem (IVT) turns this common-sense idea into a precise tool. It’s how you prove an equation has a solution without solving it, and it’s a favourite on AP free-response questions.

Key ideas

The theorem

Intermediate Value Theorem. 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 $[a, b]$ with $f(c) = k$.

In words: a continuous function takes every value between its endpoint values. When $k$ is strictly between $f(a)$ and $f(b)$, the number $c$ is strictly between $a$ and $b$.

The curve has to cross the line $y = k$ at least once to get from $f(a)$ to $f(b)$. Here it crosses three times.

What it does and doesn’t tell you

• It guarantees that at least one $c$ exists. There may be more.
• It does not tell you where $c$ is, or how many there are.
• It says nothing about values outside the range from $f(a)$ to $f(b)$. The function might reach them, or might not.

The hypothesis matters

The IVT needs $f$ to be continuous on all of $[a, b]$. A single jump or asymptote can let the function skip over $k$ (Example 3). So the first step in using the IVT is always to check continuity. Polynomials, exponentials, sine, and cosine are continuous everywhere; rational functions and logarithms are continuous on their domains (see continuity).

Finding zeros

The most common use: if $f$ is continuous on $[a, b]$ and $f(a)$ and $f(b)$ have opposite signs, then $0$ is between them, so $f(c) = 0$ for some $c$ between $a$ and $b$. To show an equation like $g(x) = h(x)$ has a solution, apply this to $f(x) = g(x) - h(x)$.

AP-style justification

Graders look for three things: continuity, the endpoint values compared to $k$, and the theorem’s name.

”$f$ is continuous on $[1, 4]$ because it is a polynomial. Since $f(1) = -2 \lt 0 \lt 3 = f(4)$, by the Intermediate Value Theorem there is a value $c$ with $1 \lt c \lt 4$ such that $f(c) = 0$.”

When $f$ is given by a table, the problem must tell you that $f$ is continuous. Say so in your answer: “Because $f$ is continuous…”

Worked examples

Example 1: Showing a root exists

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

Solution. Let $f(x) = x^3 + x - 1$. It’s a polynomial, so it’s continuous on $[0, 1]$.

$$f(0) = -1, \qquad f(1) = 1 + 1 - 1 = 1$$

Since $f(0) = -1 \lt 0 \lt 1 = f(1)$, by the Intermediate Value Theorem there is a value $c$ with $0 \lt c \lt 1$ such that $f(c) = 0$. That $c$ is a solution of the equation.

Example 2: A table of values

A continuous function $f$ has these values:

$x$	$0$	$2$	$5$	$7$
$f(x)$	$4$	$-1$	$3$	$6$

What is the fewest number of solutions $f(x) = 1$ must have on $[0, 7]$? Justify your answer.

Solution. Look for intervals where $1$ is between the endpoint values:

• On $[0, 2]$: $f(0) = 4$ and $f(2) = -1$, and $-1 \lt 1 \lt 4$. Since $f$ is continuous, the IVT gives a $c$ in $(0, 2)$ with $f(c) = 1$.
• On $[2, 5]$: $f(2) = -1$ and $f(5) = 3$, and $-1 \lt 1 \lt 3$. The IVT gives another $c$ in $(2, 5)$.
• On $[5, 7]$: $f$ goes from $3$ to $6$, and $1$ isn’t between them, so the IVT says nothing here.

The two intervals $(0, 2)$ and $(2, 5)$ don’t overlap, so $f(x) = 1$ has at least two solutions on $[0, 7]$.

Example 3: When the hypothesis fails

Let $f(x) = \dfrac{1}{x}$ on $[-1, 1]$. Then $f(-1) = -1$ and $f(1) = 1$, and $0$ is between them. Is there a $c$ in $[-1, 1]$ with $f(c) = 0$?

Solution. No: $\dfrac{1}{x}$ is never $0$. This doesn’t contradict the IVT, because $f$ is not continuous on $[-1, 1]$. It isn’t even defined at $x = 0$, where it has a vertical asymptote. The graph jumps from $-\infty$ to $\infty$ and skips over $0$.

Example 4: Solving an equation with a calculator

Show that $e^x = 3 - x$ has a solution, and find it to three decimal places.

Solution. Rewrite as $g(x) = e^x + x - 3 = 0$. The function $g$ is a sum of continuous functions, so it’s continuous everywhere. Try some easy values:

$$g(0) = 1 + 0 - 3 = -2, \qquad g(1) = e + 1 - 3 = e - 2 \approx 0.718$$

Since $g(0) \lt 0 \lt g(1)$, by the IVT there’s a $c$ in $(0, 1)$ with $g(c) = 0$, which means $e^c = 3 - c$.

The IVT doesn’t find $c$, but a graphing calculator can: graph $y = e^x + x - 3$ and find its zero, or intersect $y = e^x$ with $y = 3 - x$. The solution is $c \approx 0.792$.

Common mistakes

Forgetting to state continuity. On the AP exam, an IVT argument without “since $f$ is continuous on $[a, b]$” doesn’t get full credit, even if everything else is right.

Using the IVT on a function that isn’t continuous. Check rational functions and piecewise functions carefully. In Example 3, $f(-1)$ and $f(1)$ have opposite signs, but there’s no zero.

Claiming the IVT tells you how many solutions there are. It guarantees at least one per interval. In Example 2 the answer is “at least two”, not “exactly two”.

Applying it to a value that isn’t between f(a) and f(b). In Example 2, $1$ is not between $f(5) = 3$ and $f(7) = 6$, so the IVT says nothing on $[5, 7]$. It’s still possible that $f(x) = 1$ there; you just can’t conclude it.

Not comparing k to both endpoint values. Write the inequality, like $f(0) = -1 \lt 0 \lt 1 = f(1)$. That’s the line that shows the IVT applies.

Practice

1. (Warm-up) $f$ is continuous on $[2, 5]$, with $f(2) = 7$ and $f(5) = -1$. Must $f(c) = 0$ for some $c$ in $(2, 5)$? Must $f(c) = 8$ for some $c$?

Solution

$f(c) = 0$: yes. Since $f$ is continuous and $-1 \lt 0 \lt 7$, the IVT guarantees such a $c$.

$f(c) = 8$: not guaranteed. $8$ isn’t between $-1$ and $7$, so the IVT says nothing. $f$ might or might not reach $8$.

2. (Warm-up) Let $g(x) = \dfrac{1}{x - 3}$. Then $g(2) = -1$ and $g(4) = 1$. Does the IVT guarantee a zero of $g$ in $(2, 4)$?

Solution

No. $g$ is not continuous on $[2, 4]$, because it’s undefined at $x = 3$ (a vertical asymptote). In fact $g$ has no zeros at all.

3. (Core) Show that $\cos x = x$ has a solution in $\left[0, \tfrac{\pi}{2}\right]$ (radians), and use a calculator to find it to three decimal places.

Solution

Let $h(x) = \cos x - x$, which is continuous everywhere.

$$h(0) = 1 - 0 = 1, \qquad h\left(\frac{\pi}{2}\right) = 0 - \frac{\pi}{2} \approx -1.571$$

Since $h\left(\tfrac{\pi}{2}\right) \lt 0 \lt h(0)$, by the IVT there’s a $c$ in $\left(0, \tfrac{\pi}{2}\right)$ with $h(c) = 0$, so $\cos c = c$.

With a calculator (in radian mode), $c \approx 0.739$.

4. (Core) A particle moves along a line. Its velocity $v(t)$, in metres per second, is continuous, with these values:

$t$ (s)	$0$	$3$	$6$	$10$
$v(t)$ (m/s)	$2$	$-1$	$-3$	$4$

What is the fewest number of times $v(t) = 0$ on $0 \le t \le 10$? Justify your answer.

Solution

$v$ is continuous.

• On $[0, 3]$: $v(3) = -1 \lt 0 \lt 2 = v(0)$, so by the IVT, $v(t) = 0$ for some $t$ in $(0, 3)$.
• On $[3, 6]$: $v$ goes from $-1$ to $-3$; $0$ isn’t between them, so no conclusion.
• On $[6, 10]$: $v(6) = -3 \lt 0 \lt 4 = v(10)$, so by the IVT, $v(t) = 0$ for some $t$ in $(6, 10)$.

So $v(t) = 0$ at least twice.

5. (Core) Show that $x^5 - 3x + 1 = 0$ has at least three real solutions.

Solution

Let $f(x) = x^5 - 3x + 1$, a polynomial, so continuous everywhere.

$x$	$-2$	$-1$	$0$	$1$	$2$
$f(x)$	$-25$	$3$	$1$	$-1$	$27$

The sign changes on $[-2, -1]$, on $[0, 1]$, and on $[1, 2]$. By the IVT, $f$ has a zero in each of the intervals $(-2, -1)$, $(0, 1)$, and $(1, 2)$. These don’t overlap, so there are at least three real solutions.

6. (Core) Let $f(x) = \begin{cases} x + 1, & x \lt 1 \\ x - 3, & x \ge 1 \end{cases}$ on $[0, 2]$. Show that $f(0)$ and $f(2)$ have opposite signs, but $f(c) = 0$ has no solution in $[0, 2]$. Why doesn’t this contradict the IVT?

Solution

$f(0) = 1 \gt 0$ and $f(2) = -1 \lt 0$.

Zeros: $x + 1 = 0$ gives $x = -1$, which isn’t in the piece’s interval. $x - 3 = 0$ gives $x = 3$, which isn’t in $[0, 2]$. So there is no zero in $[0, 2]$.

There’s no contradiction, because $f$ is not continuous on $[0, 2]$: at $x = 1$, the left-hand limit is $2$ and the right-hand limit is $-2$. The hypothesis of the IVT isn’t met.

7. (Core) $f$ is continuous, with $f(1) = 4$ and $f(3) = k$. For which values of $k$ does the IVT guarantee a $c$ with $1 \lt c \lt 3$ and $f(c) = 6$?

Solution

The IVT applies when $6$ is strictly between $f(1) = 4$ and $f(3) = k$. Since $4 \lt 6$, we need $k \gt 6$.

(If $k = 6$, then $c = 3$ works, but that isn’t strictly between $1$ and $3$.)

8. (Challenge) $f$ is continuous on $[0, 1]$ and $0 \le f(x) \le 1$ for every $x$ in $[0, 1]$. Show that there is a number $c$ in $[0, 1]$ with $f(c) = c$. (Such a $c$ is called a fixed point.)

Solution

Let $g(x) = f(x) - x$. It’s continuous on $[0, 1]$ (a difference of continuous functions).

$$g(0) = f(0) - 0 \ge 0, \qquad g(1) = f(1) - 1 \le 0$$

If $g(0) = 0$, then $c = 0$ works. If $g(1) = 0$, then $c = 1$ works. Otherwise $g(1) \lt 0 \lt g(0)$, and by the IVT there’s a $c$ in $(0, 1)$ with $g(c) = 0$.

In every case $g(c) = 0$, which means $f(c) = c$.

9. (Challenge) (Calculator) Show that $\ln x = 3 - x$ has a solution between $2$ and $3$, and find it to three decimal places.

Solution

Let $g(x) = \ln x + x - 3$. It’s continuous for $x \gt 0$, so it’s continuous on $[2, 3]$.

$$g(2) = \ln 2 - 1 \approx -0.307, \qquad g(3) = \ln 3 \approx 1.099$$

Since $g(2) \lt 0 \lt g(3)$, by the IVT there’s a $c$ in $(2, 3)$ with $g(c) = 0$, so $\ln c = 3 - c$.

Using a calculator to find the zero of $g$: $c \approx 2.208$.