For example, if the temperature of the air in a room warms from $-3^\circ C$−3°C to $18^\circ C$18°C after a heater is turned on, it is not difficult to accept the proposition that there must be a moment when the temperature is exactly $0^\circ C$0°C.

This idea applies whenever there is a process of continuous change. It would not be true, however, if the cold air in the room could be suddenly replaced by warm air so that the temperature $0^\circ C$0°C is skipped.

In the long process of making the calculus, as invented by Newton and Leibniz, rigorous, the intermediate value idea had to be made explicit and it had to be given a proof. The proof depends on an understanding of continuity and on the least upper bound property of the real numbers.

Continuity

Roughly speaking, we say a function is continuous over an interval if for any small change in the domain variable there corresponds a small change in the function value. That is, there is never a jump in the function value when the domain variable changes by a tiny amount.

Least upper bound

Ordered sets, including the real numbers, have the property that whenever a set has an upper bound it must have a least upper bound.

Consider, for example, the set $\left\{x:\ x^2<2\right\}${x: x2<2}. The number $1.5$1.5 is an upper bound for this set because it is greater than any number in the set. Similarly, the number $1.45$1.45 is an upper bound because $1.45^2=2.1025>2$1.452=2.1025>2. The least upper bound property asserts that there exists an upper bound, which we call $\sqrt{2}$√2, that is less than all other upper bounds.

The theorem

The intermediate value theorem says that if $f$f is a continuous function with values $f(a)$f(a) and $f(b)$f(b) where $aa<b, then for any number $n$n between $f(a)$f(a) and $f(b)$f(b) there is a number $c$c between $a$a and $b$b such that $f(c)=n$f(c)=n.

The proof begins with the special case $f(a)<0$f(a)<0 and $f(b)>0$f(b)>0 and we show that there exists $aa<c<b such that $f(c)=0$f(c)=0.

By the continuity of $f$f, if $f(a)<0$f(a)<0, it must be possible to find numbers $w$w near $a$a such that $f(a+w)<0$f(a+w)<0. The set of all such numbers $w$w is bounded above and so it must have a least upper bound.

Call this least upper bound $c$c and consider $f(c)$f(c). The value $f(c)$f(c) cannot be negative because then $c$c would not be an upper bound. Additionally, $f(c)$f(c) cannot be positive because then $c$c would not be the least upper bound. We are left with the only remaining possibility, that $f(c)=0$f(c)=0.

To move from the special case to the more general theorem, we construct a new continuous function $g(x)=f(x)-n$g(x)=f(x)−n. Given that $f(a)f(a)<n<f(b), it follows that $f(a)-n<0f(a)−n<0<f(b)−n. That is, $g(a)<0g(a)<0<g(b).

By the previous argument, we know that there exists $c$c such that $g(c)=0$g(c)=0. So, $f(c)-n=0$f(c)−n=0 and hence, $f(c)=n$f(c)=n, as required.

The intermediate value theorem

If $f(x)$f(x) is a function that is continuous over the interval $[a,b]$[a,b], then for any number $n$n between $f(a)$f(a) and $f(b)$f(b) there is a number $c$c such that $aa<c<b and $f(c)=n$f(c)=n.

Finding zeros

One immediate implication of the intermediate value theorem is that it gives us a way to determine whether a continuous function has a zero in a given interval. Notice that if the sign of $f(a)$f(a) is different to the sign of $f(b)$f(b), then there must be some function value between $f(a)$f(a) and $f(b)$f(b) that is exactly equal to $0$0. There will then be a corresponding $x$x-value for this function value that we call the zero of the function.

Locating zeros of a function

If $f(x)$f(x) is a function that is continuous over the interval $[a,b]$[a,b], and if $f(a)$f(a) and $f(b)$f(b) have opposite signs, then there exists at least one value $c$c such that $aa<c<b and $f(c)=0$f(c)=0.

Exploration

A certain quantity varies with time according to the rule $x(t)=t^3-2t^2+1$x(t)=t3−2t2+1. It is clear that $x(1)=0$x(1)=0, but we wonder whether there are other values of $t$t such that $x(t)=0$x(t)=0.

We choose some values for $t$t. Say, $t=-1$t=−1, $t=0.5$t=0.5, $t=1.5$t=1.5, and $t=2$t=2, and we calculate $x(t)$x(t) in each case.

$t$`t`	$-1$−1	$0.5$0.5	$1.5$1.5	$2$2
$x\left(t\right)$`x`(`t`)	$-2$−2	$0.625$0.625	$-0.125$−0.125	$1$1

By the intermediate value theorem, we see that the function $x(t)$x(t) has a zero between $t=-1$t=−1 and $t=0.5$t=0.5, another between $0.5$0.5 and $1.5$1.5 (which we already knew about), and another between $t=1.5$t=1.5 and $t=2$t=2.

We conclude that there is at least three zeros in the interval between $t=-1$t=−1 and $t=2$t=2.

Practice questions

Question 1

Consider the polynomial $P\left(x\right)=4x^2-8x+2$P(x)=4x2−8x+2. $Dylan$ would like to know if it has a real zero between $x=1$x=1 and $x=2$x=2.

Find $P\left(1\right)$P(1).
Find $P\left(2\right)$P(2).
What conclusion can $Dylan$ make about $P\left(x\right)$P(x) between $x=1$x=1 and $x=2$x=2 using the intermediate value theorem?
There are no real zeros between $x=1$x=1 and $x=2$x=2.
A
$Dylan$ cannot conclude anything about the zeros of $P\left(x\right)$P(x).
B
There is exactly one real zero between $x=1$x=1 and $x=2$x=2.
C
There is at least one real zero between $x=1$x=1 and $x=2$x=2.
D

Question 2

Consider the polynomial $P\left(x\right)=4x^3-x^2+7x+7$P(x)=4x3−x2+7x+7. $Sharon$ would like to know if it has a real zero between $x=-0.8$x=−0.8 and $x=-0.7$x=−0.7.

Find $P\left(-0.8\right)$P(−0.8) to one decimal place.
Find $P\left(-0.7\right)$P(−0.7) to one decimal place.
What conclusion can $Sharon$ make about $P\left(x\right)$P(x) between $x=-0.8$x=−0.8 and $x=-0.7$x=−0.7 using the intermediate value theorem?
There is exactly one real zero between $x=-0.8$x=−0.8 and $x=-0.7$x=−0.7
A
She cannot conclude anything about the zeros of the function.
B
There is at least one real zero between $x=-0.8$x=−0.8 and $x=-0.7$x=−0.7
C
There are no real zeros between $x=-0.8$x=−0.8 and $x=-0.7$x=−0.7
D

Question 3

Consider the polynomial $P\left(x\right)=2x^3-8x^2+6x+6$P(x)=2x3−8x2+6x+6. $Yuri$ would like to know if it has a real zero between $x=2.5$x=2.5 and $x=2.6$x=2.6.

Find $P\left(2.5\right)$P(2.5) to one decimal place.
Find $P\left(2.6\right)$P(2.6) to one decimal place.
What conclusion can $Yuri$ make about $P\left(x\right)$P(x) between $x=2.5$x=2.5 and $x=2.6$x=2.6 using the intermediate value theorem?
There is exactly one real zero between $x=2.5$x=2.5 and $x=2.6$x=2.6.
A
There is at least one real zero between $x=2.5$x=2.5 and $x=2.6$x=2.6.
B
There are no real zeros between $x=2.5$x=2.5 and $x=2.6$x=2.6.
C
He cannot conclude anything about the zeros of $P\left(x\right)$P(x).
D

Intermediate Value Theorem