The Intermediate Value Theorem (IVT) talks about the values that a continuous function has to take:
| Intermediate Value Theorem: Suppose $f(x)$ is a continuous function on the interval $[a,b]$ with $f(a) \ne f(b)$. If $N$ is a number between $f(a)$ and $f(b)$, then there is a point $c$ in $(a,b)$ such that $f(c)=N$. |
In other words, to go continuously from $f(a)$ to $f(b)$, you
have to pass through $N$ along the way.
DO: Before
watching the following video, sketch a graph of a continuous
function $f$, label some $x$ values $a$ and $b$ and pick any $N$
as specified, and see if you can tell what the IVT is saying.
In this video we consider the theorem graphically and ask: What does it do for us?
We can use the IVT to show that certain equations have solutions,
or that certain polynomials have roots.
DO: Work through the
following example carefully, on your on, after watching the
video, referring to the IVT to confirm each step. Sketch
$f$ near the $a$ and $b$ values used.
Example: It is very challenging to find the roots of $f(x)=x^4+x-3$. (Try it!) However, we can use the IVT to see that it has roots: Since $f(-2)=11>0>f(0)=-3$, we can let $N=0$ and use the IVT to see that that there has to be an $x$-value $c$ between $a=-2$ and $b=0$ with $f(c)=N=0$, and thus that $c$ is a root of $f$. Likewise, since $f(0)=-3 \lt 0 \lt f(2)=15$, there has to be an $x$-value $d$ between $0$ and $2$ with $f(d)=0$. We have determined that $f(x)$ has at least two roots. We don't know exactly what these roots are, but we know they exist, and that $c$ is in the interval $(-2,0)$, and that $d$ is in the interval $(0,2)$.