Intermediate value theorem

We already know that the graph of a function will not contain a hole at any point where it is continuous, thanks to the notion of continuity at a point. The Intermediate Value Theorem states that the graph of a continuous function on a closed interval will be devoid of holes.

What is the intermediate theorem?

The continuous function theorem is the intermediate value theorem. The intermediate value theorem is essential in mathematics, especially in functional analysis. This theorem demonstrates the benefits of function continuity. The two most important instances of this theorem are frequently used in mathematics. In this post, we will study the intermediate value theorem and its two claims.

Proof of the Intermediate Value Theorem:

If f(x) is continuous on [a,b] and k is strictly between f(a) and f(b), then there exists some c in (a,b) where f(c)=k.

Proof:

Without loss of generality, let us assume that k is between f(a) and f(b) in the following way:f(a)<k<f(b). The case where f(b)<k f(a) is handled similarly.

Define a set S={x∈[a,b]:f(x)<k}, and let c be the supremum of S (i.e., the smallest value that is greater than or equal to every value of S).

Note, due to continuity at a, we can keep f(x) within any ϵ>0 off (a) by keeping x sufficiently close to a. Since f(a)<k is a strict inequality, consider the implication when ϵ is half the distance between k and f(a). No value sufficiently close to a can then be greater than, which means there are values larger than a in S. Hence, a cannot be the supremum c– some value to its right must be.

Likewise, due to the continuity at b, we can keep f(x) within any ϵ>0 off (b) by keeping x sufficiently close too. Since k<f(b) is a strict inequality, consider the similar implication when ϵ is half the distance between k and f(b). Every value sufficiently close to b must then be greater than, which means b cannot be the supremum c– some value to its left must be.

With c≠a and c≠b, but c∈[a,b], it must be the case that c is in(a,b).

Now we hope to show that f(c)=k.

Since f(x) is assumed to be continuous on[a,b] and c∈[a,b], we know that limx→c f(x)=f(c). Thus, by the epsilon-delta definition, we know that for any ϵ>0, we can find aδ>0 so that (with the exception of x=c) whenever c−δ<x<c+δ, we must have f(c)−ϵ<f(x)<f(c)+ϵ.

Consider any such value ϵ>0 and the value of δ that goes with it.

Now that there must exist some x 0∈(c−δ,c] where f(x0)<k. If there wasn’t, then c would not have been the supremum of S– some value to the left of c would have been.

However, since c−δ<x 0<c+δ, we also know that f(c)−ϵ<f(x0)<f(c)+ϵ. Note, that in the case not inferred by the continuity off(i.e., whenx=c), this just says f(c)−ϵ<f(c)<f(c)+ϵ, which should be obvious.

Focusing on the left side of this string inequality,f(c)−ϵ<f(c), we add ϵ to both sides to obtain f(c)<f(x0)+ϵ. Remembering that f(x0)<k we have

f(c)<k+ϵ(1)

There also must exist some x1∈[c,c+δ)wheref(x1)≥k. If there wasn’t, then c wo

uld not have been the supremum of S– some value to the right of c would have been.

However, since c−δ<x1<c+δ, we also know that f(c)−ϵ<f(x1)<f(c)+ϵ. (As above, the case not inferred by the continuity of is an obvious one.)

Focusing on the right side of this string inequality, f(x1)<f(c)+ϵ, we subtractϵfrom both sides to obtain f(x1)−ϵ<f(c). Remembering that f(x1)≥k we have

k−ϵ<f(c) (2)

Then combining(1)and(2), we have

k−ϵ<f(c)<k+ϵ

However, the only way this holds for anyϵ>0, is forf(c)=k.

QED.

What is the intermediate value theorem equation?

The Intermediate Value Theorem (IVT) is a precise mathematical statement (theorem) concerning the properties of continuous functions. The IVT states that if a function is continuous on [a, b], and if L is any number between f(a) and f(b),then there must be a value, x = c, where a < c < b, such that f(c) = L.

Example:

Use the intermediate value theorem to prove that the equation x^3=x+8 has at least one solution.

Solution: The function x^3=x+8 can be written as x^3-x-8=0

For f(3): (27 -(3) -8) = 16

For f(2): (8 -(2) -8) = -2

For f(0): -8

For f(-2): (-8 -(-2) -8) = -14

Since f(3) = 16, and f(2) = -2 and f(-2) = -14, there will be some value x such that f(x) = 0. And, since the function is a polynomial, it will be continuous in the interval [-2,3]. Therefore, the function x^3=x+8 does, in fact, have a real solution.

Using the Bisection method we converge on a solution by iteratively bisecting (cutting in half) an upper and lower value starting with f(-2) and f(3). Doing so, our solution is x = 2.166312754. An advanced graphing calculator such as the TI-83, 84 or 89 would be an asset in solving such problems.

Conclusion:

For a variety of reasons, the Intermediate Value Theorem is useful. First and foremost, it helps in the formation of the mathematical foundations for calculus. In fact, the IVT is a key component of the Extreme Value Theorem (EVT) and Mean Value Theorem (MVT) .