Intermediate value limit theorem proof, example

 

 

 

The intermediate value theorem illustrates that for each value connecting the least upper bound and greatest lower bound of a continuous curve, where one point lies below the line and the other point above the line, and there will be at least one place where the curve crosses the line.

Statement:
If f(x) is a continuous function on [a, b] then for every d between f (a) and f (b), there exists c between a and b so that f(c) =d.
 
Proof:
Suppose f is a continuous function on [a, b]. Let v be a real number between f (a) and f (b). Since f is continuous, it takes on every number between f (a) and f (b), i.e., every intermediate value. Thus, d is a value of f.
 
 
From the above diagram, If d is between f(a) and f(b) then there exists c ,between a and b, so that f(c)=d.
 
Let's partition [a, b] into two sets of A and B such that, each number in [a, b] is in either A or B and each number in A is less than every number in B.
 
 
Suppose if f(a)< f(b) then a parallel argument will lead to the same conclusion if
f (a)>f(b).Let α be in [a, b] such that for all x in [a, α],f(x) <= d. Let A be the set of all such α's and B be the set of points in [a, b] which are not in A, then there exists x in [a, β] such that f(x)>d. So α< x <=β. Therefore each element of A is less than every element of B.
 
Assume f(c) < d, By continuity, we have limx→d f(x) = f(c).Hence there exists δ> 0 such that
0 < |x-c| < δ => |f(x)-f(c)| < d-f(c) => f(x)-f(c) < d-f(c) => f(x)< d
 
It follows that, since (f(x) ≤ d if x is in A if x < c), f(c) < d, and (0 < |x-c| < δ if c < x ≤ c + δ/2), we obtain like this:
x ≤ c + δ/2 => f(x) < d.
 
Therefore, (c+δ/2) is in A, which means that c is in A but is not the largest number in A, which is a contradiction. So we must have f(c) ≥ d .
 
For example, assume f(c) > d, there exists δ > 0 such that:
0 < |x-c| < δ => |f(x)-f(c)| < f(c)-d => f(c)-f(x) < f(c)-d => f(x)>d
 
In particular:
c - δ < x ≤ c - δ/2 => f(x) > d.
 
Because there is x ≤ c - δ/2 such that f(x)> d, (c-δ/2) is not in A, it is in B which means that c is in B but is not the smallest number in B, which is contradiction. As a consequence, we must have f(c)≤d.
 
Hence we must have f(c) = d.
 
 
Hence the Intermediate Value Limit Theorem is proved.

Intermediate value limit theorem proof, example

The intermediate value theorem illustrates that for each value connecting the least upper bound and greatest lower bound of a continuous curve, where one point

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2024-05-08

 

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