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.