Balzano's Theorem
A particular case of the intermediate value theorem is Bolzano's theorem. This theorem proposes that if a function is continuous on a closed interval, 
and 
and 
have opposite sign then there is at least one value of x for which 
. Usually, that value of x is denoted with "c" in many textbooks.
Bolzano's theorem does not indicate the value or values of c, it only confirms their existence. Let's prove this theorem.
Bolzano's Theorem Proof
Imagine a continuous function where point a is the minimum point and point b is the maximum point,
. Now we will divide the interval into two equal parts and call the mid point c, therefore, there will be two intervals,
. There are three possibilities in this scenario and they are:
, since
interval. If
interval where
. In both conditions, there is one term a negative entity while the other term is a positive entity.
at its positive end points. We will denote this sub-interval by
where
.
The next step is to bisect the close intervals into two equal parts as shown above. This can either result in
or new sub-interval which we will name
. Where
, now we will again bisect the sub-interval.
Since,
, therefore:
This is following a sequence which is
. This process can go till infinity. Thus we get a sequence of nested intervals.
Example
Verify that the equation
has at least one real solution in the interval 
.We consider the function
, which is continuous on because it is polynomial. We study the sign in the extremes of the interval:First, consider the function
, which is continuous in because it is polynomial. Then, study the sign in the extremes of the interval:
As the signs are different, Bolzano's theorem can be applied which determines that there is a
such that . This process demonstrates that there is a solution in this interval.
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what will be sign of -infinity is to power infinity plus or minus
The sign **cannot be determined**.
For a negative base:
* **Even exponent** → **positive**
* **Odd exponent** → **negative**
So as the exponent tends to infinity, the sign depends on whether the exponent is even or odd. Therefore, ((-\infty)^\infty) has **no definite sign**—it can be **positive or negative**.