Intermediate value theorem

[EnsignParis]

Is like the most obvious thing in calculus ever.

if a function is continuous on [a,b] then there will be a value (where c=!a and c=!b (=! in this case means not equal)) and f(c) will be between f(a) and f(b).

I raised my hand in class and basically said: "so, uh, if a function is continuous between two points, then according to this, there will be a number between the two points?"

And the teacher responded with something along the lines of: "yeah, that's pretty much it."

I know it should be stated as a theorem for proof purposes, but I thought it was kinda funny. Hehe.

[Tyvek]

[Leo Wyatt]

I don't understand

[Seven of Nine]

Took me a bit, but I get it now

Yup, f(c) is a number of some sort between f(a) and f(b).

I must have missed that class in school though

[EnsignParis]

For a quick lesson:

f(x)=x+3 is the same as saying y=x+3. Whenever you see f(asymbol) = equation, you take whatever is in the parenthesis, and substitute it wherever you see the symbol in the equation. For example, using the previous function, f(x)=x+3, f(2)=2+3, or 5.

It's like this:

By the way, I actually forgot to say that the value between a and b is "c" in my original post, which may have confused some of you.

But the intermediate value theorem says that:

If a function is continuous,

Which means just what it sounds like:
a - The function has a value for every point x over the interval stated (between a and b)
b - The function has a limit at every point in said interval
c - The limit of every point in the interval=value at every point in the interval.

For example: take this function:

f(x)=[(x+3)(x-2)]/(x-2)

By factoring, we can cancel out x-2 from the numerator and the denominator, and we are left with f(x)=x+3

However, that is not what the original function was. It is (almost) equivalent to that and looks the same as that for almost every value, except when x=2. If we plugged in x=2 in the original equation like this:

f(2)=[(2+3)(2-2)]/(2-2)=5/0, which is impossible, since we cannot have a zero in the denominator (if you could divide by zero, you can actually fiddle with the numbers to make it seem that 2=1).

f(x)=[(x+3)(x-2)]/(x-2) looks just like f(x)=x+3, except we have a little hole at x=2 (Technically the hole is invisible, but to illustrate the point that x=2 does not exist in the domain of this function, we draw the hole).

This is what it looks like:



So, by our previous definition of a continuous function, this one is not continuous. It DOES have a limit at 2 (which basically means you can take any value NEAR x=2 and the closer you get to x=2, the closer you will get to f(2). However, since it does not have a value at 2, we cannot say that this is continuous on the interval of all real numbers (which is negative infinity to positive infinity), however, we CAN say that this function IS continuous on a closed interval that does not include x=2. For example, we could say that this function is continuous on the interval [8, 12] (this notation means between 8 and 12, including 8 and 12 themselves), which says that between 8 and 12, including the numbers 8 and 12 themselves, the function is continuous.

So now that we have a definition of a continuous function, the "Intermediate Value Theorem" states that

-If we have continuity on a closed interval (a,b) (this notation "(a,b)" means between a and b, but not including the values a and b themselves)
-There will be a value, "N," between f(a) and f(b) (which are the corresponding y values to a and b) (f(a) and f(b) are not equal)
-There will also be a value, "c," on the x axis, which is between a and b.
-f(c) (c's corresponding y value) will equal N.

Lets use a new function: f(x)=x+2
Let a=0 and let b=4.

So, for this function, we do have continuity on the interval of all real numbers, so on any closed interval (in this case we are using the interval between 0 and 4) we also have continuity.

Therefore, according to the intermediate value theorem, there will be an infinite amount of y values between 0 and 4, with an infinite amount of x values to correspond to these y values.

That's basically all it says.

I know I wrote a lot, and if you don't get it, don't worry. I know that math can be quite hard to read when written like that. I'd say it's much easier to show someone in person.

Anyways, have a nice day. =)