# Mathematics for Applied Sciences (Osnabrück 2023-2024)/Part I/Lecture 11

The Intermediate value theorem

We want to know what happens to an interval ${\displaystyle {}[a,b]}$ under a continuous function ${\displaystyle {}f\colon \mathbb {R} \rightarrow \mathbb {R} }$. The values ${\displaystyle {}f(a)}$ and ${\displaystyle {}f(b)}$ belong to the image. The Intermediate value theorem tells us that all numbers between ${\displaystyle {}f(a)}$ and ${\displaystyle {}f(b)}$ do also belong to the image of the interval.

## Theorem

Let ${\displaystyle {}a\leq b}$ be real numbers, and let ${\displaystyle {}f\colon [a,b]\rightarrow \mathbb {R} }$ be a continuous function. Let ${\displaystyle {}u\in \mathbb {R} }$ be a real number between ${\displaystyle {}f(a)}$ and ${\displaystyle {}f(b)}$. Then there exists an ${\displaystyle {}x\in [a,b]}$ such that ${\displaystyle {}f(x)=u}$.

### Proof

We consider the situation ${\displaystyle {}f(a)\leq u\leq f(b)}$, and show the existence of such an ${\displaystyle {}x}$ with the bisection method. For that, we put ${\displaystyle {}a_{0}:=a}$ and ${\displaystyle {}b_{0}:=b}$, we consider the arithmetic mean ${\displaystyle {}x_{0}:={\frac {a_{0}+b_{0}}{2}}}$, and we compute

${\displaystyle f{\left(x_{0}\right)}.}$

If ${\displaystyle {}f{\left(x_{0}\right)}\leq u}$, then we put

${\displaystyle a_{1}:=x_{0}\,\,{\text{ and }}\,\,b_{1}:=b_{0}}$

and if ${\displaystyle {}f{\left(x_{0}\right)}>u}$, then we put

${\displaystyle a_{1}:=a_{0}\,\,{\text{ and }}\,\,b_{1}:=x_{0}.}$

In each case, the new interval ${\displaystyle {}[a_{1},b_{1}]}$ is lying inside the initial interval and has half of its length. It fulfills again the condition ${\displaystyle {}f{\left(a_{1}\right)}\leq u\leq f{\left(b_{1}\right)}}$, therefore we can apply the same defining method again and get recursively a family of nested intervals. Let ${\displaystyle {}x}$ denote the real number which is determined by these nested intervals. For the lower bounds of the intervals, we have ${\displaystyle {}f{\left(a_{n}\right)}\leq u}$, and this carries over to the limit ${\displaystyle {}x}$, due to the criterion for continuity in terms of sequences Hence, ${\displaystyle {}f{\left(x\right)}\leq u}$. For the upper bounds, we have ${\displaystyle {}f{\left(b_{n}\right)}\geq u}$, and this again carries over to ${\displaystyle {}x}$, so ${\displaystyle {}f(x)\geq u}$.  Therefore, ${\displaystyle {}f(x)=u}$.

${\displaystyle \Box }$

The method described in this proof is constructive and can be used to give an explicit numerical method.

## Corollary

Let ${\displaystyle {}a\leq b}$ be real numbers, and let ${\displaystyle {}f\colon [a,b]\rightarrow \mathbb {R} }$ be a continuous function with ${\displaystyle {}f(a)\leq 0}$ and ${\displaystyle {}f(b)\geq 0}$. Then there exists an ${\displaystyle {}x\in [a,b]}$ such that ${\displaystyle {}f(x)=0}$.

### Proof

This follows immediately from Theorem 11.1 .

${\displaystyle \Box }$

## Method

Let ${\displaystyle {}a\leq b}$ be real numbers, and let ${\displaystyle {}f\colon [a,b]\rightarrow \mathbb {R} }$ denote a continuous function such that ${\displaystyle {}f(a)\leq 0}$ and ${\displaystyle {}f(b)\geq 0}$. Then the function has a zero within the interval, due to the Intermediate value theorem Such a zero can be found by the bisection method, as described in the proof of the Intermediate value theorem We put ${\displaystyle {}a_{0}=a}$ and ${\displaystyle {}b_{0}=b}$, and the other interval bounds are inductively defined in such a way that ${\displaystyle {}f(a_{n})\leq 0}$ and ${\displaystyle {}f(b_{n})\geq 0}$ hold. Define ${\displaystyle {}x_{n}={\frac {a_{n}+b_{n}}{2}}}$ and compute ${\displaystyle {}f(x_{n})}$. If ${\displaystyle {}f{\left(x_{n}\right)}\leq 0}$, then we set

${\displaystyle a_{n+1}:=x_{n}\,\,{\text{ and }}\,\,b_{n+1}:=b_{n},}$

and if ${\displaystyle {}f{\left(x_{n}\right)}>0}$, then we set

${\displaystyle a_{n+1}:=a_{n}\,\,{\text{ and }}\,\,b_{n+1}:=x_{n}.}$

In both cases, the new interval ${\displaystyle {}[a_{n+1},b_{n+1}]}$ has half the length of the preceding interval and so we have bisected intervals. The real number ${\displaystyle {}x}$ defined by these nested intervals is a zero of the function.

## Example

We want to determine approximately a zero for the polynomial

${\displaystyle {}f(x)=x^{3}-4x+2\,}$

with the help of the method given by the Intermediate value theorem. We have ${\displaystyle {}f(1)=-1}$ and ${\displaystyle {}f(2)=2}$, hence by Corollary 11.2 , there must be a zero inside the interval ${\displaystyle {}[1,2]}$. We compute the value of the function at the arithmetic mean of the interval, which is ${\displaystyle {}{\frac {3}{2}}}$, and get

${\displaystyle {}f{\left({\frac {3}{2}}\right)}={\frac {27}{8}}-4\cdot {\frac {3}{2}}+2={\frac {27-48+16}{8}}={\frac {-5}{8}}<0\,.}$

Hence, we have to continue with the right half of the interval ${\displaystyle {}[{\frac {3}{2}},2]}$. The arithmetic mean thereof is ${\displaystyle {}{\frac {7}{4}}}$. The value of the function at this point is

${\displaystyle {}f{\left({\frac {7}{4}}\right)}={\left({\frac {7}{4}}\right)}^{3}-4\cdot {\frac {7}{4}}+2={\frac {343}{64}}-5={\frac {343-320}{64}}={\frac {23}{64}}>0\,.}$

So now we have to continue with the left part of the interval, namely ${\displaystyle {}[{\frac {3}{2}},{\frac {7}{4}}]}$. Its arithmetic mean is ${\displaystyle {}{\frac {13}{8}}}$. The value of the function at this point is

${\displaystyle {}f{\left({\frac {13}{8}}\right)}={\left({\frac {13}{8}}\right)}^{3}-4\cdot {\frac {13}{8}}+2={\frac {2197}{512}}-{\frac {13}{2}}+2={\frac {2197-3328+1024}{512}}={\frac {-107}{512}}<0\,.}$

Therefore, we know that there is a zero between ${\displaystyle {}{\frac {13}{8}}}$ and ${\displaystyle {}{\frac {7}{4}}={\frac {14}{8}}}$.

## Remark

The existence of arbitrary roots of nonnegative real numbers follows also from the Intermediate value theorem since, for ${\displaystyle {}c\geq 0}$, the continuous function ${\displaystyle {}X^{k}-c}$ attains negative and positive values and has therefore also a zero. The proof of Theorem 8.13 rests on the method of the Intermediate value theorem, even though continuity is not explicitly used.

## Example

Suppose that a regular quadratic table with four table legs ${\displaystyle {}A,B,C,D}$ is standing on an uneven but stepfree underground. At the moment, it stands on the legs ${\displaystyle {}A,B,C}$, and the leg ${\displaystyle {}D}$ does not touch the ground (if we leave ${\displaystyle {}B,C}$ in their positions and put ${\displaystyle {}D}$ down to the ground, then ${\displaystyle {}A}$ would sink into the ground). We claim that we can bring the table, by turning it around its middle axis, into a position such that it stands on all four legs (we do not claim that the table is then horizontal). To see this, we consider the function which assigns to the angle of rotation, the height of ${\displaystyle {}D}$ above the ground, when the three other legs are (or would be) on the ground. This height might be negative (if we are on sand, this can be realized, else think of this "ideally“). In degree ${\displaystyle {}0}$, the height is positive. In degree ${\displaystyle {}90}$, we get a situation which is symmetric to the initial position, but still the legs ${\displaystyle {}A,B,C}$ are supposed to be on the ground. Hence, the height of ${\displaystyle {}D}$ is now negative. The function has, on the interval ${\displaystyle {}[0,90]}$, positive and also negative values. Since the function is continuous, by the Intermediate value theorem it also has a zero.

Continuous bijective functions and their inverse functions

For a bijective continuous function on a real interval, its inverse function is again continuous.

## Theorem

Let ${\displaystyle {}I\subseteq \mathbb {R} }$ be an interval and let

${\displaystyle f\colon I\longrightarrow \mathbb {R} }$

denote a continuous, strictly increasing function. Then the image

${\displaystyle {}J:=f(I)\,}$

is also an interval, and the inverse function

${\displaystyle f^{-1}\colon J\longrightarrow I}$

is also continuous.

### Proof

This proof was not presented in the lecture.
${\displaystyle \Box }$

Root functions

## Theorem

Let ${\displaystyle {}n\in \mathbb {N} _{+}}$. For ${\displaystyle {}n}$ odd, the power function

${\displaystyle \mathbb {R} \longrightarrow \mathbb {R} ,x\longmapsto x^{n},}$
${\displaystyle \mathbb {R} \longrightarrow \mathbb {R} ,x\longmapsto x^{1/n},}$

is strictly increasing and continuous. For ${\displaystyle {}n}$ even, the power function

${\displaystyle \mathbb {R} _{\geq 0}\longrightarrow \mathbb {R} _{\geq 0},x\longmapsto x^{n},}$

is continuous strictly increasing, bijective, and the inverse function

${\displaystyle \mathbb {R} _{\geq 0}\longrightarrow \mathbb {R} _{\geq 0},x\longmapsto x^{1/n},}$

is strictly increasing and continuous.

### Proof

The continuity follows from Corollary 10.7 . The increasing behavior for ${\displaystyle {}x\geq 0}$ follows from the Binomial theorem For ${\displaystyle {}n}$ odd and ${\displaystyle {}x<0}$, the growth behavior follows from ${\displaystyle {}x^{n}=-(-x)^{n}}$, and the behavior in positive part. The growth behavior gives injectivity. For ${\displaystyle {}x\geq 1}$, we have ${\displaystyle {}x^{n}\geq x}$, which shows that there is no upper bound for the image. For ${\displaystyle {}n}$ odd, we get that there is no lower bound for the image. Due to the Intermediate value theorem the image is ${\displaystyle {}\mathbb {R} }$ or ${\displaystyle {}\mathbb {R} _{\geq 0}}$. Hence, the power functions are surjective, and the inverse functions exist. The continuity of the inverse functions follows from Theorem 11.7 .

${\displaystyle \Box }$

## Example

The speed of sound on earth depends on the temperature. If we work with the absolute temperature ${\displaystyle {}T}$ (measured in Kelvin), then the relation ${\displaystyle {}v=20,06{\sqrt {T}}}$ holds, where the speed of sound is measured in ${\displaystyle {}m/s}$. For ${\displaystyle {}T=300K}$, the speed of sound is about ${\displaystyle {}347,5m/s}$.

The theorem of Bolzano-Weierstraß

The following statement is called theorem of Bolzano-Weierstraß.

## Theorem

Let ${\displaystyle {}{\left(x_{n}\right)}_{n\in \mathbb {N} }}$ denote a bounded sequence of real numbers. Then this sequence has a convergent subsequence.

### Proof

Suppose that the sequence ${\displaystyle {}{\left(x_{n}\right)}_{n\in \mathbb {N} }}$ is bounded by

${\displaystyle {}a_{0}\leq x_{n}\leq b_{0}\,.}$

We define inductively an interval bisection, such that in all the intervals, there are infinitely many members of the sequence. The initial interval is ${\displaystyle {}I_{0}:=[a_{0},b_{0}]}$. Suppose that the ${\displaystyle {}k}$-th interval ${\displaystyle {}I_{k}}$ is already constructed. We consider the two halves

${\displaystyle [a_{k},{\frac {a_{k}+b_{k}}{2}}]\,\,{\text{ and }}\,\,[{\frac {a_{k}+b_{k}}{2}},b_{k}].}$

Al least in one of these, there are infinitely many members of the sequence, and we choose the interval ${\displaystyle {}I_{k+1}}$ as one half with infinitely many members. By this method, the lengths of the intervals are bisected, and so we have a sequence of nested intervals. As a subsequence, we choose arbitrary elements

${\displaystyle {}x_{n_{k}}\in I_{k}\,}$

with ${\displaystyle {}n_{k}>n_{k-1}}$. This is possible, as each interval contains infinitely members. This subsequence converges due to Exercise 8.21 to the number ${\displaystyle {}x}$ determined by the nested intervals

${\displaystyle \Box }$

Minima and maxima

## Definition

Let ${\displaystyle {}M}$ denote a set, and

${\displaystyle f\colon M\longrightarrow \mathbb {R} }$

a function. We say that ${\displaystyle {}f}$ attains in a point ${\displaystyle {}x\in M}$ its maximum, if

${\displaystyle f(x)\geq f(x'){\text{ holds for all }}x'\in M,}$

and that ${\displaystyle {}f}$ attains in ${\displaystyle {}x}$ its minimum, if

${\displaystyle f(x)\leq f(x'){\text{ holds for all }}x'\in M.}$

The common name for a maximum or a minimum is extremum. In the preceding definition, we also talk about the global maximum, since the property refers to all elements of the domain of the definition. When we are only interested in the behavior on an open, maybe small, neighborhood, then the concept of a local maximum is relevant.

## Definition

Suppose that ${\displaystyle {}D\subseteq \mathbb {R} }$ is a subset and let

${\displaystyle f\colon D\longrightarrow \mathbb {R} }$

denote a function. We say that ${\displaystyle {}f}$ attains a local maximum in a point ${\displaystyle {}x\in D}$, if there exists some ${\displaystyle {}\epsilon >0}$, such that for all ${\displaystyle {}x'\in D}$ fulfilling ${\displaystyle {}\vert {x-x'}\vert \leq \epsilon }$, the estimate

${\displaystyle {}f(x)\geq f(x')\,}$

holds. We say that ${\displaystyle {}f}$ attains a local minimum in a point ${\displaystyle {}x\in D}$, if there exists some ${\displaystyle {}\epsilon >0}$, such that for all ${\displaystyle {}x'\in D}$ fulfilling ${\displaystyle {}\vert {x-x'}\vert \leq \epsilon }$, the estimate

${\displaystyle {}f(x)\leq f(x')\,}$
holds.

If ${\displaystyle {}f(x)>f(x')}$ holds for all ${\displaystyle {}x'\neq x}$, then we talk about an isolated maximum.

## Theorem

Let ${\displaystyle {}[a,b]\subseteq \mathbb {R} }$ be a closed bounded interval, and let

${\displaystyle f\colon [a,b]\longrightarrow \mathbb {R} }$

denote a continuous function. Then there exists some ${\displaystyle {}x\in [a,b]}$ such that

${\displaystyle f(x)\geq f(x'){\text{ for all }}x'\in [a,b].}$
This means that the function attains its maximum

(and its minimum).

### Proof

This proof was not presented in the lecture.
${\displaystyle \Box }$

With the help of differential calculus, we will get to know effective methods to determine minima and maxima.

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