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# MTH5P2C: LOCATION OF ROOTS

The location of roots theorem is one of the most intutively obvious properties of continuous functions, as it states that if a continuous function attains positive and negative values, it must have a root (i.e. it must pass through 0).

## Statement

Let $f:[a,b]\rightarrow\mathbb{R}$ be a continuous function such that $f(a)<0$ and $f(b)>0$. Then there is some $c\in (a,b)$ such that $f(c)=0$.

## Proof

Let $A=\{x|x\in [a,b],\; f(x)<0\}$

As $a\in A$ $A$ is non-empty. Also, as $A\subset [a,b]$ $A$ is bounded

Thus $A$ has a least upper bound, $\sup A = u \in A.$

If $f(u)<0$:

As $f$ is continuous at $u$ $\exists\delta>0$ such that $x\in V_{\delta}(u)\implies f(x)<0$, which contradicts (1).

Also if $f(u)>0$: $f$ is continuous imples $\exists\delta>0$ such that $x\in V_{\delta}(u)\implies f(x)>0$, which again contradicts (1) by the Gap lemma.

Hence, $f(u)=0$.

The assertion of the Intermediate Value Theorem is something which is probably ‘intuitively obvious’, and is also provably true: if a function f

f

is continuous on an interval [a,b]

[a,b]

and if f(a)<0

f(a)<0

and f(b)>0

f(b)>0

(or vice-versa), then there is some third point c

c

with a<c<b

a<c<b

so that f(c)=0

f(c)=0

. This result has many relatively ‘theoretical’ uses, but for our purposes can be used to give a crude but simple way to locate the roots of functions. There is a lot of guessing, or trial-and-error, involved here, but that is fair. Again, in this situation, it is to our advantage if we are reasonably proficient in using a calculator to do simple tasks like evaluating polynomials! If this approach to estimating roots is taken to its logical conclusion, it is called the method of interval bisection, for a reason we’ll see below. We will not pursue this method very far, because there are better methods to use once we have invoked thisjust to get going.

#### Example 1

For example, we probably don’t know a formula to solve the cubic equation

x3x+1=0

x3−x+1=0

But the function f(x)=x3x+1

f(x)=x3−x+1

is certainly continuous, so we can invoke the Intermediate Value Theorem as much as we’d like. For example, f(2)=7>0

f(2)=7>0

and f(2)=5<0

f(−2)=−5<0

, so we know that there is a root in the interval [2,2]

[−2,2]

. We’d like to cut down the size of the interval, so we look at what happens at the midpoint, bisecting the interval [2,2]

[−2,2]

: we have f(0)=1>0

f(0)=1>0

. Therefore, since f(2)=5<0

f(−2)=−5<0

, we can conclude that there is a root in [2,0]

[−2,0]

. Since both f(0)>0

f(0)>0

and f(2)>0

f(2)>0

, we can’t say anything at this point about whether or not there are roots in [0,2]

[0,2]

. Again bisecting the interval [2,0]

[−2,0]

where we know there is a root, we compute f(1)=1>0

f(−1)=1>0

. Thus, since f(2)<0

f(−2)<0

, we know that there is a root in [2,1]

[−2,−1]

(and have no information about [1,0]

[−1,0]

).

If we continue with this method, we can obtain as good an approximation as we want! But there are faster ways to get a really good approximation, as we’ll see.

Unless a person has an amazing intuition for polynomials (or whatever), there is really no way to anticipate what guess is better than any other in getting started.

#### Example 2

Invoke the Intermediate Value Theorem to find an interval of length 1

1

or less in which there is a root of x3+x+3=0

x3+x+3=0

: Let f(x)=x3+x+3

f(x)=x3+x+3

. Just, guessing, we compute f(0)=3>0

f(0)=3>0

. Realizing that the x3

x3

term probably ‘dominates’ f

f

when x

x

is large positive or large negative, and since we want to find a point where f

f

is negative, our next guess will be a ‘large’ negative number: how about 1

−1

? Well, f(1)=1>0

f(−1)=1>0

, so evidently 1

−1

is not negative enough. How about 2

−2

? Well, f(2)=7<0

f(−2)=−7<0

, so we have succeeded. Further, the failed guess 1

−1

actually was worthwhile, since now we know that f(2)<0

f(−2)<0

and f(1)>0

f(−1)>0

. Then, invoking the Intermediate Value Theorem, there is a root in the interval [2,1]

[−2,−1]

.

Of course, typically polynomials have several roots, but the number of roots of a polynomial is never more than its degree. We can use the Intermediate Value Theorem to get an idea where all of them are.

#### Example 3

Invoke the Intermediate Value Theorem to find three different intervals of length 1

1

or less in each of which there is a root of x34x+1=0

x3−4x+1=0

: first, just starting anywhere, f(0)=1>0

f(0)=1>0

. Next, f(1)=2<0

f(1)=−2<0

. So, since f(0)>0

f(0)>0

and f(1)<0

f(1)<0

, there is at least one root in [0,1]

[0,1]

, by the Intermediate Value Theorem. Next, f(2)=1>0

f(2)=1>0

. So, with some luck here, since f(1)<0

f(1)<0

and f(2)>0

f(2)>0

, by the Intermediate Value Theorem there is a root in [1,2]

[1,2]

. Now if we somehow imagine that there is a negative root as well, then we try 1

−1

f(1)=4>0

f(−1)=4>0

. So we know nothing about roots in [1,0]

[−1,0]

. But continue: f(2)=1>0

f(−2)=1>0

, and still no new conclusion. Continue: f(3)=14<0

f(−3)=−14<0

. Aha! So since f(3)<0

f(−3)<0

and f(2)>0

f(2)>0

, by the Intermediate Value Theorem there is a third root in the interval [3,2]

[−3,−2]