# Intermediate Value Theorem

The idea behind the Intermediate Value Theorem is this:

When we have

**two points**connected by a continuous curve:

one point below the line

the other point above the line

then there is **at least one place** where the curve crosses the line!

Well **of course** we must cross the line to get from A to B!

Now that you know the **idea**, let's look more closely at the details.

## Continuous

The curve must be **continuous** ... no gaps or jumps in it.

Continuous is a special term with an exact definition in calculus, but here we will use this simplified definition:

we can draw it without lifting our pen from the paper

## More Formal

Here is the Intermediate Value Theorem stated more formally:

When:

- The curve is the function y = f(x),
- which is continuous on the interval [a, b],
- and w is a number between f(a) and f(b),

Then ...

... there must be at least one value **c** within [a, b] such that **f(c) = w**

In other words the function y = f(x) at some point must be w = f(c)

Notice that:

**w**is between f(a) and f(b),*which leads to ...***c**must be between a and b

## At Least One

It also says "at least one value c", which means we **could** have more.

Here, for example, are 3 points where f(x)=w:

## How Is This Useful?

Whenever we can show that:

- there is a point above some line
- and a point below that line, and
- that the curve is continuous,

we can then safely say "yes, there is a value somewhere **in between** that is on the line".

### Example: is there a solution to x^{5} − 2x^{3} − 2 = 0
between x=0 and x=2?

**At x=0:**

0^{5} − 2 × 0^{3} − 2 = −**2**

**At x=2:**

2^{5 }− 2 × 2^{3} − 2 = **14**

Now we know:

- at x=0, the curve is below zero
- at x=2, the curve is above zero

And, being a polynomial, the curve will be continuous,

so **somewhere in between** the curve must cross through y=0

Yes, there is a solution to x^{5} − 2x^{3} − 2 = 0
in the interval [0, 2]

## An Interesting Thing!

### The Intermediate Value Theorem Can Fix a Wobbly Table

If your table is wobbly because of uneven ground ... ... just The ground must be continuous (no steps such as poorly laid tiles). |

### Why does this work?

We can always have 3 legs on the ground, it is the 4th leg that is the trouble.

Imagine we are **rotating the table**, and the 4th leg could somehow go into the ground (like sand):

- at some point it will be above the ground
- at another point it will be below the ground

So there must be some point where the 4th leg **perfectly touches the ground** and the table won't wobble.

*(The famous Martin Gardner wrote about this in Scientific American. There is also a very complicated proof somewhere).*

## Another One

At some point during a round-trip you will be

exactly as high
as where you started.

(It only works if you don't start at the highest or lowest point.)

The idea is:

- at some point you will be higher than where you started
- at another point you will be lower than where you started

So there must be a point in between where you are **exactly** as high as where you started.

Oh, and your path must be **continuous**, no disappearing and reappearing somewhere else.

The same thing happens with temperature, pressure, and so on.

## And There's More!

If you follow a circular path ... somewhere on that circle there will be points that are:

- directly opposite each other
**and**at the same height!

two points that are

directly opposite **and** at same height

*Can you think of more examples?*