intermediate value theorem

 

In Mathematical analysis, the intermediate value theorem is either of two theorems of which an account is given below.

Intermediate value theorem

The intermediate value theorem states the following: If y=f(x) is continuous on [a,b], and N is a number between f(a) and f(b), then there is at least 1 c ∈ [a,b] such that f(c) = N.

Suppose that I is an interval [a,b] in the real numbers R and that is a continuous function. Then the image set Failed to parse (unknown function\displaystyle): \displaystyle f(I)

is also an interval, and either it contains [f(a),f(b)], or it contains [f(b),f(a)]; that is,

• Failed to parse (unknown function\displaystyle): \displaystyle f(I) \supseteq [f(a),f(b)]

, or

• Failed to parse (unknown function\displaystyle): \displaystyle f(I) \supseteq [f(b),f(a)]

.

It is frequently stated in the following equivalent form: Suppose that is continuous and that u is a real number satisfying f(a) < u < f(b) or f(a) > u > f(b). Then for some c ∈ [a,b], f(c) = u.

This captures an intuitive property of continuous functions: given f continuous on [1,2], if f(1) = 3 and f(2) = 5 then f must be equal to 4 somewhere between 1 and 2. It represents the idea that the graph of a continuous function can be drawn without lifting your pencil from the paper.

The theorem depends on the completeness of the real numbers. It is false for the rational numbers Q. For example, the function f(x) = x^2 - 2, x ∈ Q satisfies . However there is no rational number Failed to parse (unknown function\displaystyle): \displaystyle x

such that Failed to parse (unknown function\displaystyle): \displaystyle f(x) = 0

.

Proof

We shall prove the first case Failed to parse (unknown function\displaystyle): \displaystyle f(a) < u < f (b)

the second is similar.

Let Failed to parse (unknown function\displaystyle): \displaystyle \text{S} = \{ x \in [a,b] : f(x) \leq u \} . Then Failed to parse (unknown function\displaystyle): \displaystyle S

is non-empty (since Failed to parse (unknown function\displaystyle): \displaystyle a \in \displaystyle S

) and bounded above by Failed to parse (unknown function\displaystyle): \displaystyle b . Hence by the completeness property of the real numbers, the supremum Failed to parse (unknown function\text): c = \sup \text{S}

exists. We claim that Failed to parse (unknown function\displaystyle): \displaystyle f(c) = u

.

Suppose first that Failed to parse (unknown function\displaystyle): \displaystyle f(c) > u . Then Failed to parse (unknown function\displaystyle): \displaystyle f(c) - u > 0 , so there is a Failed to parse (unknown function\displaystyle): \displaystyle \delta > 0

such that Failed to parse (unknown function\displaystyle): \displaystyle |f(x) - f(c)| < f(c) - u
whenever Failed to parse (unknown function\displaystyle): \displaystyle |x - c| < \delta

, since Failed to parse (unknown function\displaystyle): \displaystyle f

is continuous. But then Failed to parse (unknown function\displaystyle): \displaystyle f(x) > f(c) - (f(c) - u) = u
whenever Failed to parse (unknown function\displaystyle): \displaystyle |x - c| < \delta
(i.e. Failed to parse (unknown function\displaystyle): \displaystyle f(x) > u
for Failed to parse (unknown function\displaystyle): \displaystyle x
in Failed to parse (unknown function\displaystyle): \displaystyle(c - \delta, c + \delta)

). Thus Failed to parse (unknown function\displaystyle): \displaystyle c - \delta

is an upper bound for Failed to parse (unknown function\displaystyle): \displaystyle S

, a contradiction since we assumed that Failed to parse (unknown function\displaystyle): \displaystyle c

was the least upper bound and Failed to parse (unknown function\displaystyle): \displaystyle c - \delta < \displaystyle c

.

Suppose next that Failed to parse (unknown function\displaystyle): \displaystyle f(c) < u . Again, by continuity, there is a Failed to parse (unknown function\displaystyle): \displaystyle \delta > 0

such that Failed to parse (unknown function\displaystyle): \displaystyle |f(x) - f(c)| < u - f(c)
whenever Failed to parse (unknown function\displaystyle): \displaystyle |x - c| < \delta

. Then Failed to parse (unknown function\displaystyle): \displaystyle f(x) < f(c) + (u - f(c)) = u

for Failed to parse (unknown function\displaystyle): \displaystyle x
in Failed to parse (unknown function\displaystyle): \displaystyle(c - \delta, c + \delta)
and there are numbers Failed to parse (unknown function\displaystyle): \displaystyle x
greater than Failed to parse (unknown function\displaystyle): \displaystyle c
for which Failed to parse (unknown function\displaystyle): \displaystyle f(x) < u

, again a contradiction to the definition of Failed to parse (unknown function\displaystyle): \displaystyle c .

We deduce that Failed to parse (unknown function\displaystyle): \displaystyle f(c) = u

as stated.

History

For Failed to parse (unknown function\displaystyle): \displaystyle u = 0

above, the statement is also known as Bolzano's theorem; this theorem was first stated by Bernard Bolzano, together with a proof which used techniques which were especially rigorous for their time but which are now regarded as non-rigorous.

Generalization

The intermediate value theorem can be seen as a consequence of the following two statements from topology:

• If Failed to parse (unknown function\displaystyle): \displaystyle X

and Failed to parse (unknown function\displaystyle): \displaystyle Y
are topological spaces, f:X→Y is continuous, and Failed to parse (unknown function\displaystyle): \displaystyle X
is connected, then Failed to parse (unknown function\displaystyle): \displaystyle f(X)
is connected.

• A subset of is connected if and only if it is an interval.

Example of use in proof

The theorem is rarely applied with concrete values; instead, it gives some characterization of continuous functions. For example, let Failed to parse (unknown function\displaystyle): \displaystyle g(x) = f(x) - x

for Failed to parse (unknown function\displaystyle): \displaystyle f
continuous over the real numbers.  Also, let Failed to parse (unknown function\displaystyle): \displaystyle f
be bounded (above and below).  Then we can say Failed to parse (unknown function\displaystyle): \displaystyle g = 0
at least once.  To see this, consider the following:

Since Failed to parse (unknown function\displaystyle): \displaystyle f

is bounded, we can pick  and .  Clearly Failed to parse (unknown function\displaystyle): \displaystyle g(a) < 0
and Failed to parse (unknown function\displaystyle): \displaystyle g(b) > 0

. If Failed to parse (unknown function\displaystyle): \displaystyle f

is continuous, then Failed to parse (unknown function\displaystyle): \displaystyle g
is also continuous. Since Failed to parse (unknown function\displaystyle): \displaystyle g
is continuous, we can apply the intermediate value theorem and state that Failed to parse (unknown function\displaystyle): \displaystyle g
must take on the value of 0 somewhere between Failed to parse (unknown function\displaystyle): \displaystyle a
and Failed to parse (unknown function\displaystyle):  \displaystyle b

. This result proves that any continuous bounded function must cross the function, Failed to parse (unknown function\displaystyle): \displaystyle x .

Converse is false

Suppose Failed to parse (unknown function\displaystyle): \displaystyle f

is a real-valued function defined on some interval Failed to parse (unknown function\displaystyle): \displaystyle I

, and for every two elements Failed to parse (unknown function\displaystyle): \displaystyle a

and Failed to parse (unknown function\displaystyle): \displaystyle b
in Failed to parse (unknown function\displaystyle): \displaystyle I
and  such that Failed to parse (unknown function\displaystyle): \displaystyle f(c) = u

. Does Failed to parse (unknown function\displaystyle): \displaystyle f

have to be continuous? The answer is no; the converse of the intermediate value theorem fails. As an example, take the function Failed to parse (unknown function\displaystyle): \displaystyle f(x) = \sin(1/x)
for x≠0, and Failed to parse (unknown function\displaystyle): \displaystyle f(0) = 0

. This function is not continuous as the limit when Failed to parse (unknown function\displaystyle): \displaystyle x

gets close to 0 does not exist; yet the function has the above intermediate value property.  Another, more complicated example is given by the Conway base 13 function.

Historically, this intermediate value property has been suggested as a definition for continuity of real-valued functions; this definition was not adopted.

Darboux's theorem states that all functions that result from the differentiation of some other function on some interval have the intermediate value property (even though they need not be continuous).

Implications of theorem in real world

The theorem implies that on any great circle around the world, the temperature, pressure, elevation, carbon dioxide concentration, or anything else that varies continuously, there will always exist two antipodal points that share the same value for that variable.

Proof: Take Failed to parse (unknown function\displaystyle): \displaystyle f

to be any continuous function on a circle. Draw a line through the center of the circle, intersecting it at two opposite points Failed to parse (unknown function\displaystyle): \displaystyle A
and Failed to parse (unknown function\displaystyle): \displaystyle B

. Let Failed to parse (unknown function\displaystyle): \displaystyle d

be the difference Failed to parse (unknown function\displaystyle): \displaystyle f(A)-f(B)

. If the line is rotated 180 degrees, the value Failed to parse (unknown function\displaystyle): \displaystyle -d

will be obtained instead. Due to the intermediate value theorem there must be some intermediate rotation angle for which Failed to parse (unknown function\displaystyle): \displaystyle d=0

, and as a consequence Failed to parse (unknown function\displaystyle): \displaystyle f(A)=f(B)

at this angle.

This is a special case of a more general result called the Borsuk–Ulam theorem.

The theorem also underpins the explanation of why rotating a wobbly table will bring it to stability (subject to certain easily-met constraints)[1]

Intermediate value theorem of integration

The intermediate value theorem of integration is derived from the mean value theorem and states:

If Failed to parse (unknown function\displaystyle): \displaystyle f

is a continuous function on some interval Failed to parse (unknown function\displaystyle): \displaystyle [a,b]

, then the signed area under the function on that interval is equal to the length of the interval Failed to parse (unknown function\displaystyle): \displaystyle b-a

multiplied by some function value Failed to parse (unknown function\displaystyle): \displaystyle f(c)
such that Failed to parse (unknown function\displaystyle): \displaystyle a < c < b

. I.e.,

Failed to parse (unknown function\displaystyle): \displaystyle \int_{a}^{b}\! f(x)\,dx = (b-a)f(c).

Intermediate value theorem of derivatives

If Failed to parse (unknown function\displaystyle): \displaystyle f

is a differentiable real-valued function on , then the (first order) derivative Failed to parse (unknown function\displaystyle): \displaystyle f'
has the intermediate value property, though Failed to parse (unknown function\displaystyle): \displaystyle f'
might not be continuous.

External links

• Intermediate value Theorem - Bolzano Theorem at cut-the-knot

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