Class IX

Intermediate Value Theorem

Lesson

The intermediate value theorem makes definite an idea that may seem intuitively obvious if we rely on everyday experience.

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

Outcomes

9.A.P.1

Definition of a polynomial in one variable, its coefficients, with examples and counterexamples, its terms, zero polynomial. Degree of a polynomial. Constant, linear, quadratic, cubic polynomials; monomials, binomials, trinomials. Factors and multiples. Zeros/roots of a polynomial/equation.