The Intermediate Value Theorem

A continuous graph can't get from height

f(a)

to height

f(b)

without crossing every height in between. When

N

is strictly between

f(a)

and

f(b)

, the number

c

is strictly between

a

and

b

. The theorem says only that such a

c

exists. It doesn't say where, or how many there are.

A continuous f meets the height N three times between a and b.

With a jump, the graph can skip N: no x in [a, b] gives the height N.

The second picture shows why the theorem needs continuity. The function starts below

N

and ends above it, but the jump carries it past

N

without taking that value.

Locating a root

Example 1

Show that the equation

x^3 + x - 5 = 0

has a solution between

1

and

2

. Then use the midpoint

1.5

to find a shorter interval that contains one.

Let

f(x) = x^3 + x - 5

. Then

f(1) = -3

and

f(2) = 5

. The justification an AP reader expects names the hypothesis, the values, and the theorem:

Since $f$ is a polynomial, it's continuous on $[1, 2]$ . Since $f(1) = -3 < 0 < 5 = f(2)$ , the Intermediate Value Theorem gives a number $c$ in $(1, 2)$ with $f(c) = 0$ .

To narrow it, check the midpoint:

f(1.5) = 3.375 + 1.5 - 5 = -0.125

. That's negative, while

f(2)

is positive, so the same argument on

[1.5, 2]

shows that this shorter interval contains a solution.

Reading a table

Example 3

The function

g

is continuous on

[0, 6]

, and selected values are given in the table.

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

(a) Find the least number of solutions of

g(x) = x

[0, 6]

(b) Explain whether

g(c) = 4

must hold for some

c

(1, 3)

g(c) = 4

must hold for some

c

(0, 1)

(a) Let

h(x) = g(x) - x

, which is continuous on

[0, 6]

. Its values at the table's inputs are

$x$	$0$	$1$	$3$	$4$	$6$
$h(x)$	$2$	$-1$	$2$	$-1$	$1$

The sign of

h

changes on each of

[0, 1]

[1, 3]

[3, 4]

, and

[4, 6]

. By the Intermediate Value Theorem on each one,

h

has a zero strictly inside it, so

g(x) = x

has at least

4

solutions in

[0, 6]

. The least number is

4

: joining the table's points with line segments gives a continuous

g

with exactly four.

(b) Yes. Since

g(1) = 0 < 4 < 5 = g(3)

and

g

is continuous on

[1, 3]

, the Intermediate Value Theorem gives a

c

(1, 3)

with

g(c) = 4

(c) Not necessarily. The values

g(0) = 2

and

g(1) = 0

don't have

4

between them, so the theorem says nothing. The graph of

g

could rise above

4

between

0

and

1

, or it could stay at or below

2

When the theorem says nothing

Example 4

Let

f(x) = \frac{1}{x}

. Then

f(-1) = -1

and

f(1) = 1

, and

0

is between them. Explain why there's no

c

[-1, 1]

with

f(c) = 0

, and why this doesn't contradict the Intermediate Value Theorem.

The equation

\frac{1}{x} = 0

has no solution at all, since a fraction with numerator

1

is never

0

. The theorem doesn't apply, because

f

isn't continuous on

[-1, 1]

. It isn't even defined at

0

, where it has an infinite discontinuity. The graph never crosses the x-axis. It gets from height

-1

to height

1

by dropping to

-\infty

and coming back from

\infty

on the other side of the asymptote

x = 0

y = 1/x on [−1, 1]: negative on the left, positive on the right, and never 0.