Polynomials

Hong Kong

Stage 4 - Stage 5

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.

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.

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`: `x`2<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 intermediate value theorem says that if $f$`f` is a continuous function with values $f(a)$`f`(`a`) and $f(b)$`f`(`b`) where $a`a`<`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 $a`a`<`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<0`f`(`a`)−`n`<0<`f`(`b`)−`n`. That is, $g(a)<0`g`(`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 $a`a`<`c`<`b` and $f(c)=n$`f`(`c`)=`n`.

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 $a`a`<`c`<`b` and $f(c)=0$`f`(`c`)=0.

A certain quantity varies with time according to the rule $x(t)=t^3-2t^2+1$`x`(`t`)=`t`3−2`t`2+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.

Consider the polynomial $P\left(x\right)=4x^2-8x+2$`P`(`x`)=4`x`2−8`x`+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.ADylan cannot conclude anything about the zeros of $P\left(x\right)$

`P`(`x`).BThere is exactly one real zero between $x=1$

`x`=1 and $x=2$`x`=2.CThere is at least one real zero between $x=1$

`x`=1 and $x=2$`x`=2.D

Consider the polynomial $P\left(x\right)=4x^3-x^2+7x+7$`P`(`x`)=4`x`3−`x`2+7`x`+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`=−0.7AShe cannot conclude anything about the zeros of the function.

BThere is at least one real zero between $x=-0.8$

`x`=−0.8 and $x=-0.7$`x`=−0.7CThere are no real zeros between $x=-0.8$

`x`=−0.8 and $x=-0.7$`x`=−0.7D

Consider the polynomial $P\left(x\right)=2x^3-8x^2+6x+6$`P`(`x`)=2`x`3−8`x`2+6`x`+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.AThere is at least one real zero between $x=2.5$

`x`=2.5 and $x=2.6$`x`=2.6.BThere are no real zeros between $x=2.5$

`x`=2.5 and $x=2.6$`x`=2.6.CHe cannot conclude anything about the zeros of $P\left(x\right)$

`P`(`x`).D