Continuity and the Intermediate Value Theorem

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Continuity and the Intermediate Value Theorem

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The Intermediate Value Theorem

If a function

$f$ is continuous at every point

$a$ in an interval

$I$ , we'll say that

$f$ is continuous on

$I$ .

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 $(0,2)$ .