## Intermediate Value Theorem – Limits and Continuity

Posted: 12th February 2013 by seanmathmodelguy in Lectures

#### Intermediate Value Theorem

To begin with, let’s start with the basic statement of the theorem.

###### Theorem

If $$f(x)$$ is continuous on a closed interval $$[a,b]$$ and $$N$$ is any number $$f(a) < N < f(b)$$ then there exists a value $$c \in (a,b)$$ such $$f(c) = N$$.

The illustration corresponding to the theorem is to the right and indicates that there may be more than one possible value for $$c$$. The important restrictions are that

• $$f(x)$$ be continuous and
• the interval $$[a,b]$$ is closed.

The primary purpose of this theorem is to indicate when numbers with various properties exist.

###### Steps

1. Make sure the function, $$f(x)$$ is continuous.
2. Create a new function $$g(x)=f(x)-N$$, replacing the function in this manner always makes the $$N$$ in the theorem with respect to $$g$$ equal to zero. So that $$g(c)=0$$ when a correct value for $$c$$ is determined.
3. Using 0 rather than the general $$N$$, we need to find an $$a$$ and a $$b$$ so that either $$g(a)>0$$ and $$g(b)<0$$ or $$g(a)<0$$ and $$g(b)>0$$. The point is that the signs need to change.
4. Finding a change of sign confirms that there is a number $$c \in (a,b)$$ that allows $$g(c)=0$$ or $$f(c)=N$$.

###### Examples

A. Suppose we have the function $$f(x) = x^2 – 4x$$ and we wish to show there is a number $$x_*$$ such that $$f(x_*) = 1$$.
1. Notice that since $$f(x)$$ is continuous, the intermediate value theorem can be used.
2. Let $$g(x) = f(x) – 1 = x^2 – 4x – 1$$ so that $$g(x) = 0$$ when the correct $$x_*$$ is determined.
3. Choosing $$a = 4$$ gives $$g(a) = -1 < 0$$ and choosing $$b=5$$ gives $$g(b) = 4 > 0$$. There are of course many other possible value of $$a$$ and $$b$$. Note that $$a<b$$.
4. Since a change in sign was found, there is a number $$c \in (4,5)$$ such that $$g(c) = 0$$ or equivalently, $$f(c) = 1$$.

B. If $$f(x) = x^3-8x+10$$, show there is at least one value of $$c$$ for which $$f(c) = -\sqrt{3}$$.

Since $$f(x)$$ is continuous we just need to redefine the function (to make $$N = 0$$) and find values for $$a$$ and $$b$$. The new function is
$g(x) = f(x) + \sqrt{3} = x^3-8x+10+\sqrt{3}.$ We need to find $$a$$ and $$b$$ so that $$g(x)$$ changes sign. Let $$a=-4$$ so that $$f(a) = -22+\sqrt{3} < 0$$ and $$b=-3$$ so that $$f(b) = 7+\sqrt{3} >0$$. These choices for $$a$$ and $$b$$ are found by just trying different values in the function.

At any rate, using the intermediate value theorem we can conclude that there is a value $$c \in (-4,-3)$$ such that $$g(c) = 0$$ or $$f(c) = -\sqrt{3}$$.

1. Kiru Sengal says:

Hi Sean,
I ran across this while scooping UOIT’s Math grad website.

I’m not so convinced about the -N traslation of the function f and it’s purpose. The IVT is basically equivalent to Dedekind completeness (or LUB property ) of the reals. Your reasoning “work” for over the rationals too (because in a sense, they look “complete” when you graph functions over the rationals). And the -N seems unnecessary too.

Thanks