# Math Insight

### Intermediate Value Theorem, location of roots

The assertion of the Intermediate Value Theorem is something which is probably ‘intuitively obvious’, and is also provably true: if a function $f$ is continuous on an interval $[a,b]$ and if $f(a) < 0$ and $f(b) > 0$ (or vice-versa), then there is some third point $c$ with $a < c < b$ so that $f(c)=0$. 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 this just to get going.

#### Example 1

For example, we probably don't know a formula to solve the cubic equation $$x^3-x+1=0$$ But the function $f(x)=x^3-x+1$ is certainly continuous, so we can invoke the Intermediate Value Theorem as much as we'd like. For example, $f(2)=7 > 0$ and $f(-2)=-5 < 0$, so we know that there is a root in the interval $[-2,2]$. We'd like to cut down the size of the interval, so we look at what happens at the midpoint, bisecting the interval $[-2,2]$: we have $f(0)=1 > 0$. Therefore, since $f(-2)=-5 < 0$, we can conclude that there is a root in $[-2,0]$. Since both $f(0) > 0$ and $f(2) > 0$, we can't say anything at this point about whether or not there are roots in $[0,2]$. Again bisecting the interval $[-2,0]$ where we know there is a root, we compute $f(-1)=1 > 0$. Thus, since $f(-2) < 0$, we know that there is a root in $[-2,-1]$ (and have no information about $[-1,0]$).

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 $1$ or less in which there is a root of $x^3+x+3 =0$: Let $f(x)=x^3+x+3$. Just, guessing, we compute $f(0)=3 > 0$. Realizing that the $x^3$ term probably ‘dominates’ $f$ when $x$ is large positive or large negative, and since we want to find a point where $f$ is negative, our next guess will be a ‘large’ negative number: how about $-1$? Well, $f(-1)=1 > 0$, so evidently $-1$ is not negative enough. How about $-2$? Well, $f(-2)=-7 < 0$, so we have succeeded. Further, the failed guess $-1$ actually was worthwhile, since now we know that $f(-2) < 0$ and $f(-1) > 0$. Then, invoking the Intermediate Value Theorem, there is a root in the interval $[-2,-1]$.

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 $1$ or less in each of which there is a root of $x^3-4x+1=0$: first, just starting anywhere, $f(0)=1 > 0$. Next, $f(1)=-2 < 0$. So, since $f(0) > 0$ and $f(1) < 0$, there is at least one root in $[0,1]$, by the Intermediate Value Theorem. Next, $f(2)=1 > 0$. So, with some luck here, since $f(1) < 0$ and $f(2) > 0$, by the Intermediate Value Theorem there is a root in $[1,2]$. Now if we somehow imagine that there is a negative root as well, then we try $-1$: $f(-1)=4 > 0$. So we know nothing about roots in $[-1,0]$. But continue: $f(-2)=1 > 0$, and still no new conclusion. Continue: $f(-3)=-14 < 0$. Aha! So since $f(-3) < 0$ and $f(2) > 0$, by the Intermediate Value Theorem there is a third root in the interval $[-3,-2]$.

Notice how even the ‘bad’ guesses were not entirely wasted.