Intermediate Value Theorem (IVT)

Here's another statement that seems so obvious it hardly needs justification: any continuous function, moving from one value to another, must pass through every real value between them. More precisely, if we have $f(x)$ continuous on $[a,b]$, then for every $c$ on the closed interval between $f(a)$ and $f(b)$, there must be a $x_0 \in [a,b]$ such that $f(x_0) = c$. This may hardly seem worth mentioning- just look at the graph below!

How could the function possibly get from the first point to the second, while staying continuous, and not cross the intermediate line? The proof, however, is not so obvious. It is, however, completely optional, so skip over what follows if you like!

Let $A$ be the set of all $x \in [a,b]$ such that $f(x) \leq c$, and let $B$ be the set of all $x \in [a,b]$ so that $f(x) > c$. We claim that $A$ contains the desired value. Let $x_0$ be the supremum or least upper bound of $A$. Note that as $A$ is closed, $x_0 \in A$. We claim that $f(x_0) = c$. Now as $f$ is continuous at $x_0$, if we choose an $\epsilon > 0$, there must be a $\delta >0$ so that if $|x-x_0| < \delta$, $|f(x) - f(x_0)| < \epsilon$. We can rewrite this conclusion as $f(x) - \epsilon < f(x_0) < f(x) + \epsilon$.

Now, since $x_0$ is the supremum, there must be values of $x \in A$ such that $x_0-\delta < x < x_0$. For these $x$, we have $f(x_0) < f(x) + \epsilon \leq c+\epsilon$. Similarly, no $x \in (x_0,x_0 + \delta)$ is in $A$, so for these we have $c-\epsilon < f(x)-\epsilon < f(x_0)$. Thus, for all $\epsilon > 0$, we have $c-\epsilon<f(x_0)<c+\epsilon$. Clearly $f(x_0) = c$.

Continuous Function and IVT Statement

This is a basic theorem in calculus. To know what it means, we first need to know what a continuous function is. We will assume you know what a limit is.

A function $f$ is a continuous at $x=a$ if $a$ is in domain of the function $f$, exists and $\lim\limits_{x\to a}f(x)=f(a)$. Intuitively, this just means that there is no gap in the graph of $f(x)$ at $x=a$ (hence the name continuous).

If a function is continuous at every point $x$ of the domain, then the function is continuous. The following theorem is easy to prove. An interval $[a,b]$ means it contains all the numbers $x\in\mathbb R$ where $a\leq x\leq b$.

Theorem If $f$ is continuous in the interval $[a,b]$ then there is a $c$ so that $f(c)$ is maximum i.e. $a\leq c\leq b$ with $f(c)\geq f(x)$ for all $x\in[a,b]$. Similarly, there is a $c$ such that $f(c)$ is minimum.

Now, the following theorem should be intuitively obvious too.

Theorem (Intermediate Value Theorem) Let $f$ be a continuous function in the interval $[a,b]$. Then, for any $w$ between $f(a)$ and $f(b)$ (both for $f(a)\leq w\leq f(b)$ and $f(a)\geq w\geq f(b)$) there is a $c$ such that $f(c)=w$.

Since $f$ is continuous, it must pass through $f(c)$ at some point. That would be our $c$. Hope you can make sense of the theorem. Note that, such a point $c$ may not be unique!

(Video) Intermediate Value Theorem

by PatrickJMT

Here's a video by patrickJMT explaing the Intermediate Value Theorem.

The Intermediate Value Theorem states: suppose you have a continuous function on a closed interval

If you take any point in between $M$ and $N$, let's call it $I$, there has to be a point in the interval $[a,b]$ that when you plug it into your function you get $I$ back out.

Suppose $f(x)$ is continuous on $[a,b]$; let $f(a)=M$, $f(b)=N$; let $M\leq I\leq N$ $\Rightarrow$ $\exists$ $c$ in $[a,b]$ such that $f(c)=I$

The firs example he looks at is to show that there is a root for $x^3-3x+1=0$ on the interval $(0,1)$.

So what we want to do is plug the end values into the function. $$f(0)=0^3-3(0)+1=1\\ f(1)=1^3-3(1)+1=-1$$ So if you start above the $x$-axis and end below the $x$-axis, then the Intermediate Value Theorem says that there's at least one point in our function that's on the $x$-axis. Remember that points on the $x$-axis are the roots.

So by the Intermediate Value Theorem we can state that there is at least one root to this function on this interval. There might be more than one but there's definitely at least one.

Try it Yourself!

Show that there is a root for $x^2=\sqrt{x+1}$ on the interval $(1,2)$.