Intermediate value theorem – Serlo

The intermediate value theorem says that every continuous function $f:[a,b]\to \mathbb {R}$ attains every value between $f(a)$ and $f(b)$ at least once. Continuous functions reach every intermediate value between $f(a)$ and $f(b)$ (if there are no holes in the domain between $a$ and $b$). So the intermediate value theorem can be used to determine the existence of functional values.

Motivation

Intermediate value theorem:: Intuition and practical application (in German)

Let $f:[a,b]\to \mathbb {R}$ be an arbitrary continuous function. At the point $a$, it has the functional value $f(a)$ and at the point $b$ it has the functional value $f(b)$. Let us assume that $f(a)\leq f(b)$. Furthermore, let $s$ be an arbitrary value between $f(a)$ and $f(b)$, meaning $f(a)\leq s\leq f(b)$:

The function values f(a) and f(b) at the points a and b, as well as the intermediate value s

By our current perception, continuous functions don't have any jumps in their domain. Since $f$ is defined on the entire interval $[a,b]$ and its domain is connected, the graph binds the points $(a,f(a))$ and $(b,f(b))$ without jumps. If we bind $f(a)$ and $f(b)$ without "taking our pencil off the paper," at some point we must cross the line $y=s$. So there is at least one point of intersection between the line $y=s$ and the graph of $f$:

The graph of f must intersect the line y=s at least once. There, one has f(x)=s

For the $x$-values ${\tilde {x}}$ of the points of intersection it holds $f({\tilde {x}})=s$. The intermediate value $s$ is therefore attained once by the function $f$. We have intuitively seen that continuous functions attain all values between any two functional values at least once, provided the domain doesn't contain any gaps between the two arguments.

The Intermediate Value Theorem

Explanation and definition of the intermediate value theorem. (YouTube video (in German) by the channel Quatematik)

Theorem (Intermediate Value Theorem)

Let $f:[a,b]\to \mathbb {R}$ be a continuous function with $a,b\in \mathbb {R}$ and $a<b$. Let $s$ be some value between the two functional values $f(a)$ and $f(b)$. This means it holds $f(a)\leq s\leq f(b)$ or $f(b)\leq s\leq f(a)$. Then there exists at least one real number $x\in [a,b]$ with $f(x)=s$. The intermediate value $s$ is attained at least once by the function $f$.

Bolzano's Root Theorem

For the proof of the intermediate value theorem, it suffices to prove this for the special case $s=0$. This special case is also called "Bolzano's Root Theorem."

Theorem (Bolzano's Root Theorem)

Let $h:[a,b]\to \mathbb {R}$ be a continuous function with $a,b\in \mathbb {R}$ and $a<b$. Furthermore let zero be an intermediate value of $h(a)$ and $h(b)$, meaning $h(a)\leq 0\leq h(b)$ or $h(a)\geq 0\geq h(b)$. Then $h$ has at least one root. This means, there exists at least one argument ${\tilde {x}}\in [a,b]$ such that $h({\tilde {x}})=0$.

Why is it sufficient to only consider this special case? Let's take a function $f:[a,b]\to \mathbb {R}$ and a value $s\in \mathbb {R}$ between the functional values $f(a)$ and $f(b)$. By the intermediate value theorem, we have to find a ${\tilde {x}}\in [a,b]$ with $f({\tilde {x}})=s$. Now it holds:

$f({\tilde {x}})=s\iff f({\tilde {x}})-s=0$

Therefore $f({\tilde {x}})=s$ if and only if $f({\tilde {x}})-s=0$. Now we will define the helping function $h:[a,b]\to \mathbb {R}$ with $h(x)=f(x)-s$. As we've already determined, the equation $h({\tilde {x}})=0$ is satisfied precisely in the case $f({\tilde {x}})=s$. So if we find a root of $h$, then the function $f$ also attains the value $s$.

Now the function $h$ fulfills all of the requirements to use Bolzano's Root Theorem. It's a function of the form $[a,b]\to \mathbb {R}$ with the closed interval $[a,b]$ as its domain. As a concatenation of continuous functions, the function $h$ is continuous. In the case $f(a)\leq s\leq f(b)$ it holds:

${\begin{aligned}f(a)\leq s\iff f(a)-s\leq 0\iff h(a)\leq 0\\[0.3em]f(b)\geq s\iff f(b)-s\geq 0\iff h(b)\geq 0\end{aligned}}$

Using $f(a)\leq s\leq f(b)$ we can deduce the chain of inequalities $h(a)\leq 0\leq h(b)$. Now if we consider the case $f(a)\geq s\geq f(b)$:

${\begin{aligned}f(a)\geq s\iff f(a)-s\geq 0\iff h(a)\geq 0\\[0.3em]f(b)\leq s\iff f(b)-s\leq 0\iff h(b)\leq 0\end{aligned}}$

Altogether we can concude that zero is an intermediate value of $h(a)$ and $h(b)$. Therefore $h$ satisfies the requirements for Bolzano's Root Theorem. By this root theorem there exists a ${\tilde {x}}\in [a,b]$ with $h({\tilde {x}})=0$. For this ${\tilde {x}}$ it holds $f({\tilde {x}})=s$. This shows that the general intermediate value theorem can be easily deduced from Bolzano's root theorem. So now we must only prove Bolzano's theorem.

Proof of Bolzano's Root Theorem

Proof (Bolzano's Root Theorem)

Let $h:[a,b]\to \mathbb {R}$ be a continuous function with $h(a)\leq 0\leq h(b)$ or $h(a)\geq 0\geq h(b)$. In the following we will consider the case $h(a)\leq 0\leq h(b)$. The other case $h(a)\geq 0\geq h(b)$ can be proved similarly. We have to find a root of $h$. This can be shown by finding a fitting sequence of nested intervals (this means that any interval in the sequence is a subset of the preceding interval in the sequence). As a starting interval we choose $[a_{1},b_{1}]=[a,b]$, so $a_{1}=a$ and $b_{1}=b$. Namely, we know that the desired root must lie in the interval $[a,b]$.

For $h(a_{1})=0$ or $h(b_{1})=0$ we already have found a root at $x=a_{1}$ or $x=b_{1}$, respectively. In this case, we are already finished. In the case $h(a_{1})<0$ and $h(b_{1})>0$, we reduced the size of our interval. For this purpose we consider the midpoint ${\tfrac {a_{1}+b_{1}}{2}}$ of the starting interval. If the value of $h$ at this point is equal to zero, then we've found a root and we are finished.

If $h\left({\tfrac {a_{1}+b_{1}}{2}}\right)\neq 0$, then we choose another interval which must contain a root. Let us assume that $h\left({\tfrac {a_{1}+b_{1}}{2}}\right)>0$. Then we get the following image:

Illustration: if the function value is positive in the middle.

We can see that the graph must cross the $x$-axis from $a_{1}$ to ${\tfrac {a_{1}+b_{1}}{2}}$. Since $h$ is a continuous function and can't have any jumps, and since this interval doesn't contain any gaps, there must be a root of $h$ in this interval. Since both functional values $h\left({\tfrac {a_{1}+b_{1}}{2}}\right)$ and $h(b_{1})$ are positive, we can't say whether this root lies in ${\tfrac {a_{1}+b_{1}}{2}}$ to $b_{1}$. For this reason we choose a second interval $[a_{2},b_{2}]=\left[a_{1},{\tfrac {a_{1}+b_{1}}{2}}\right]$.

If $h\left({\tfrac {a_{1}+b_{1}}{2}}\right)$ is less than zero, the graph of $h$ has to undergo a change in sign in the second interval from ${\tfrac {a_{1}+b_{1}}{2}}$ to $h(b_{1})$. Correspondingly, a zero should lie in this interval and in this case we choose $\left[{\tfrac {a_{1}+b_{1}}{2}},b_{1}\right]$ as the second interval $[a_{2},b_{2}]$:

Illustration: if the function value is negative in the middle.

Altogether we determine $[a_{2},b_{2}]$ in the following way:

$[a_{2},b_{2}]={\begin{cases}\left[a_{1},{\frac {a_{1}+b_{1}}{2}}\right]&;h\left({\frac {a_{1}+b_{1}}{2}}\right)>0\\\left[{\frac {a_{1}+b_{1}}{2}},b_{1}\right]&;h\left({\frac {a_{1}+b_{1}}{2}}\right)<0\end{cases}}$

We keep repeating this process: in the $n$-th step we calculate the midpoint ${\tfrac {a_{n}+b_{n}}{2}}$ of the interval $[a_{n},b_{n}]$. If $h$ attains the value $0$ here, we are finished and can return ${\tfrac {a_{n}+b_{n}}{2}}$ as a root. Otherwise we choose a new interval $[a_{n+1},b_{n+1}]$ using the following definition:

$[a_{n+1},b_{n+1}]={\begin{cases}\left[a_{n},{\frac {a_{n}+b_{n}}{2}}\right]&;h\left({\frac {a_{n}+b_{n}}{2}}\right)>0\\\left[{\frac {a_{n}+b_{n}}{2}},b_{n}\right]&;h\left({\frac {a_{n}+b_{n}}{2}}\right)<0\end{cases}}$

By this procedure we obtain either the desired root after a certain iteration $n$ or we get a sequence of intervals $([a_{n},b_{n}])_{n\in \mathbb {N} }$. By this method of choosing the sequence elements, the sequence $(a_{n})_{n\in \mathbb {N} }$ is monotone increasing and the sequence $(b_{n})_{n\in \mathbb {N} }$ is monotone falling. Since every sequential element lies in the interval $[a,b]$, the sequences are both bounded. Then we can use Monotoniekriterium für Folgen to conclude that both sequences converge. Note that the length of each interval is cut in half at each iteration, meaning:

${\begin{aligned}\lim _{n\rightarrow \infty }{(b_{n}-a_{n})}&=\lim _{n\rightarrow \infty }{{\frac {1}{2}}\cdot (b_{n-1}-a_{n-1})}\\[0.3em]&=\lim _{n\rightarrow \infty }{{\frac {1}{2}}\cdot {\frac {1}{2}}\cdot (b_{n-2}-a_{n-2})}\\[0.3em]&\vdots \\[0.3em]&=\lim _{n\rightarrow \infty }{{\frac {1}{2^{n}}}\cdot (b_{1}-a_{1})}=0\end{aligned}}$

From here it follows $\lim _{n\rightarrow \infty }{a_{n}}=\lim _{n\rightarrow \infty }{b_{n}}$. For this reason, the sequences of lower and upper interval limits converge to the same value $c\in [a;b]$. Furthermore, by our construction, it holds that $h(a_{n})<0$ and $h(b_{n})>0$ for all $n\in \mathbb {N}$. For this reason it holds for the limits: $\lim _{n\rightarrow \infty }{h(a_{n})}\leq 0$ and $\lim _{n\rightarrow \infty }{h(b_{n})}\geq 0$. Because $h$ is continuous, it holds

$\lim _{n\rightarrow \infty }{h(a_{n})}=h\left(\lim _{n\rightarrow \infty }{a_{n}}\right)=h(c)=h\left(\lim _{n\rightarrow \infty }{b_{n}}\right)=\lim _{n\rightarrow \infty }{h(b_{n})}$

From the above line we can conclude $h(c)\leq 0$ and $h(c)\geq 0$, and therefore $h(c)=0$ must hold. In this case we've also found a root of the function $h$.

Corollaries of the Intermediate Value Theorem

Continuous functions map intervals to intervals

With the help of the intermediate value theorem we can prove that continuous functions map intervals to intervals.

Theorem (Corollary of the Intermediate Value Theorem)

Let $I\subseteq \mathbb {R}$ be an interval and $f:I\to \mathbb {R}$ a continuous function. Then $f(I)$ is also an interval.

Proof (Corollary of the Intermediate Value Theorem)

We set $A=\inf f(I)$ and $B=\sup f(I)$. We also allow $A=-\infty$ (if $f(I)$ is bounded from below) and $B=\infty$ (if $f(I)$ is bounded from above). We now take some real number $y$ with $A<y<B$. From the definition of the infimum and the fact that $A<y$ we get that there exists some $a\in I$ with $f(a)<y$. Similarly there exists some $b\in I$ with $y<f(b)$. Altogether we have $f(a)<y<f(b)$.

So $y$ is an intermediate value of two functional values of $f$. Since $f$ is continuous, the intermediate value theorem guarantees us the existence of some $x\in I$ with $f(x)=y$. Since $y\in (A,B)$ was arbitrary, it holds $(A,B)\subseteq f(I)$. Now $A$ or $B$ could be elements of $f(I)$. Therefore $f(I)$ must be one of the following four intervals:

${\begin{aligned}f(I)={\begin{cases}[A,B]&A{\text{ and }}B{\text{ are attained by }}f\\[0.5em][A,B)&A{\text{ is attained by }}f{\text{ , but not }}B\\[0.5em](A,B]&B{\text{ is attained by }}f{\text{ , but not }}A\\[0.5em](A,B)&A{\text{ and }}B{\text{ are not attained by }}f\end{cases}}\end{aligned}}$

We see that $f(I)$ is an interval.

Range of Power Functions

Example (Range of power functions $x\mapsto x^{k}$)

The power functions $f:\mathbb {R} \to \mathbb {R}$ with $f(x)=x^{k}$ for $k\in \mathbb {N}$ are continuous. Furthermore it holds

$\lim _{x\to \infty }f(x)=\lim _{x\to \infty }x^{k}=\infty$

Damit ist $\sup f(\mathbb {R} )=\infty$. Außerdem ist

$\lim _{x\to -\infty }f(x)=\lim _{x\to -\infty }x^{k}={\begin{cases}\infty &{\text{ for even }}k\\-\infty &{\text{ for odd }}k\end{cases}}$

For odd $k$ it holds $\inf f(\mathbb {R} )=-\infty$. For even $k$ it holds $\inf f(\mathbb {R} )=0$, since for all $x\in \mathbb {R}$ it holds $f(x)=x^{k}\geq 0$ and zero is attained as a value because $f(0)=0^{k}=0$. The by the intermediate value theorem we have

$f(\mathbb {R} )={\begin{cases}\mathbb {R} _{0}^{+}=[0,\infty )&{\text{ for even }}k\\\mathbb {R} =(-\infty ,\infty )&{\text{ for odd }}k\end{cases}}$

Range of the Exponential Function

Example (Range of the exponential function)

The exponential function $\exp :\mathbb {R} \to \mathbb {R}$ is continuous. Also it holds $\exp(x)>0$ for all $x\in \mathbb {R}$. It holds $\lim _{x\to -\infty }\exp(x)=0$ and $\lim _{x\to \infty }\exp(x)=\infty$. Then $\inf \exp(\mathbb {R} )=0$ and $\sup \exp(\mathbb {R} )=\infty$. Since neither $0$ nor $\infty$ are attained by the exponential function, the intermediate value theorem says

$\exp(\mathbb {R} )=\mathbb {R} ^{+}=(0,\infty )$

Exercise (Domain of the Generalized Exponential Function)

For $a\in \mathbb {R} \setminus \{1\}$ ,

$\exp _{a}:\mathbb {R} \to \mathbb {R} ,\ \exp _{a}(x)=a^{x}=\exp(x\ln(a))$

defines the generalized exponential function. Show: $\exp _{a}(\mathbb {R} )=\mathbb {R} ^{+}$.

Tip: Differentiate between the cases $a>1$ and $0<a<1$.

Solution (Domain of the Generalized Exponential Function)

Since $\exp$ and $x\mapsto x\cdot \ln(a)$ are continuous on $\mathbb {R}$, it holds that $\exp _{a}$ is also continuous (as the concatenation of continuous functions) on all of $\mathbb {R}$. Further, $\exp _{a}(x)>0$, since $\exp(x)>0$.

Fall 1: $a>1$

Here it holds $\ln a>0$, and therefore $\lim _{x\to \infty }x\ln a=\infty$ holds. From here it follows $\lim _{x\to \infty }\exp _{a}(x)=\lim _{x\to \infty }\exp(x\ln a)=\infty$. On the other hand, $\lim _{x\to -\infty }x\ln a=-\infty$, and therefore $\lim _{x\to -\infty }\exp _{a}(x)=\lim _{x\to -\infty }\exp(x\ln a)=0$. Altogether this case yields $\exp _{a}(\mathbb {R} )=(0,\infty )=\mathbb {R} ^{+}$.

Fall 2: $0<a<1$

Here it holds $\ln a<0$, and therefore $\lim _{x\to \infty }x\ln a=-\infty$. We then can conclude $\lim _{x\to \infty }\exp _{a}(x)=\lim _{x\to \infty }\exp(x\ln a)=0$. On the other hand it holds $\lim _{x\to -\infty }x\ln a=\infty$, and therefore $\lim _{x\to -\infty }\exp _{a}(x)=\lim _{x\to -\infty }\exp(x\ln a)=\infty$. So it also holds in this case $\exp _{a}(\mathbb {R} )=(0,\infty )=\mathbb {R} ^{+}$.

Exercise: Fixed-point theorem

Proof of a Fixed-point Theorem

In the following exercise we will prove a fixed-point theorem. Fixed points are arguments $x$ of a function $f$ that satisfy the equation $f(x)=x$. In a sentence: fixed points are those points not changed by a function transformation. Fixed point theorems are therefore theorems, that prove the existence of fixed points in certain situations. For mathematics, such theorems are important because sometimes we can reduce the problem of proving the existence of a certain object to proving the existence of a fixed point. For example, the argument $x$ is a root of the function $f:\mathbb {R} \to \mathbb {R}$, if and only if the function $g:\mathbb {R} \to \mathbb {R}$ with the ordering $g(x)=x-f(x)$ has a fixed point. Using the existence of a fixed point of the function $g$, we can prove the existence of a root for $f$. In the following exercise we are going to prove a kind of intermediate value theorem:

Exercise (Fixed point theorem)

Let $f:[a,b]\to [a,b]$ be a continuous function. Show that $f$ has at least one fixed point. Fixed points are points $x\in [a,b]$ with $f(x)=x$.

How to get to the proof? (Fixed point theorem)

By rearranging the equation $f(x)=x$ we obtain $f(x)-x=0$. Therefore $x$ is a fixed point of $f$ if and only if $x$ is a root of the function $h(x)=f(x)-x$. So let us define a helping function $h:[a,b]\to \mathbb {R}$ with $h(x)=f(x)-x$. As you might have guessed, we can use Bolzano's Root Theorem here to prove the existence of a root. For that we have to prove that $h$ satisfies all the assumptions of the theorem:

• $h$ is continuous
• Zero is an intermediate value of $h(a)$ and $h(b)$.

$h$ is continuous as the concatenation of continuous functions and we can also prove that $h(b)\leq 0\leq h(a)$ holds. The root theorem guarantees us the existence of the limit.

Proof (Fixed point theorem)

Let $f:[a,b]\to [a,b]$ be a continuous function. We define the helping function $h:[a,b]\to \mathbb {R} :x\mapsto h(x)=f(x)-x$. For this function it holds:

• $h$ is continuous on $[a,b]$ as the difference of the continuous functions $f$ and ${\text{id}}:[a,b]\to [a,b]:x\mapsto x$.
• It holds $h(b)\leq 0\leq h(a)$ because:
  • $h(a)=\underbrace {f(a)} _{\in [a,b]}-a\geq 0$
  • $h(b)=\underbrace {f(b)} _{\in [a,b]}-b\leq 0$.

Therefore $h$ fulfills the assumptions of the root theorem. Then there exists some ${\tilde {x}}\in [a,b]$ with $h({\tilde {x}})=f({\tilde {x}})-{\tilde {x}}=0$. This means $f({\tilde {x}})={\tilde {x}}$. In other words, $f$ has a fixed point.

Assumptions of the Fixed Point Theorem

In the fixed point theorem above, the continuity is a necessary condition of the proven theorem. If we leave this condition out, we can find a function $g:[a,b]\to [a,b]$ for which this theorem is no longer true. This will be shown in the following exercise:

Exercise (Fixed Point Theorem)

Find a function $g:[0,1]\to [0,1]$ with $g(x)\neq x$ for all $x\in [0,1]$.

Proof (Fixed Point Theorem)

For example, the function

$g:[0,1]\to [0,1]:x\mapsto g(x)={\begin{cases}1&{\text{ für }}0\leq x\leq {\tfrac {1}{2}},\\0&{\text{ für }}{\tfrac {1}{2}}<x\leq 1\end{cases}}$

doesn't have a fixed point. This is shown in the following graph of the function $g$. Namely, there is not point of intersection of the graph of $g$ with the graph of the identity function ${\text{id}}:[0,1]\to [0,1],\ x\mapsto x$:

Graph of the function g and of the identity function f

Exercise: Roots and Range of Polynomials

Roots of Polynomials

The following exercise exemplifies a special case of the fundamental theorem of algebra. This theorem says that a non-constant polynomial

$p(z)=a_{n}z^{n}+a_{n-1}z^{n-1}+\ldots +a_{1}z+a_{0}$

with complex coefficients $a_{0},a_{1},\ldots ,a_{n-1},a_{n}$ has at least one complex root. In the real case this claim does not hold in general. A polynomial with real coefficients need not necessarily have real roots. A polyomial function without real roots could be, for example, $p(x)=x^{2}+1$. For certain polynomials, we can still prove the existence of a root:

Exercise (Roots of Polynomials)

Let

$p:\mathbb {R} \to \mathbb {R} ,\ p(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\ldots +a_{1}x+a_{0}$

be a polynomial function with $a_{0},a_{1},\ldots ,a_{n-1},a_{n}\in \mathbb {R}$ and $a_{n}\neq 0$. If $n$, meaning if the degree of $p$ is odd, then $p$ has at least one (real) root.

Summary of proof (Roots of Polynomials)

We will prove the exercise with the help of the root theorem. For that, we need to find two real numbers$x_{1},x_{2}\in \mathbb {R}$ with $p(x_{1})\leq 0\leq p(x_{2})$ or $p(x_{1})\geq 0\geq p(x_{2})$. This is a consequence of $\lim _{x\to \infty }p(x)=\infty$ and $\lim _{x\to -\infty }p(x)=-\infty$, which we can show by using a clever rewriting of $p$ for the case $a_{n}>0$. We will now thoroughly investigate the case $a_{n}>0$. The case $a_{n}<0$ can be handles similarly.

Solution (Roots of Polynomials)

We will demonstrate the proof for $a_{n}>0$. The case $a_{n}<0$ is similar. For $x\neq 0$ we can factor out $x^{n}$ to obtain:

$p(x)=x^{n}(a_{n}+a_{n-1}x^{-1}+\ldots +a_{1}x^{-n+1}+a_{0}x^{-n})$

The expression $a_{n}+a_{n-1}x^{-1}+\ldots +a_{1}x^{-n+1}+a_{0}x^{-n}$ in the parentheses converges for $x\to \pm \infty$ to $a_{n}$. This holds since the terms $x^{-1},\ldots ,x^{-n}$ converge for $x\to \pm \infty$ to $0$. Since $n$ is odd, it holds:

$\lim _{x\to \infty }p(x)=\lim _{x\to \infty }\underbrace {x^{n}} _{\to \infty }(\underbrace {a_{n}+a_{n-1}x^{-1}+\ldots +a_{1}x^{-n+1}+a_{0}x^{-n}} _{\to a_{n}>0})=\infty$

und

$\lim _{x\to -\infty }p(x)=\lim _{x\to -\infty }\underbrace {x^{n}} _{\to -\infty }(\underbrace {a_{n}+a_{n-1}x^{-1}+\ldots +a_{1}x^{-n+1}+a_{0}x^{-n}} _{\to a_{n}>0})=-\infty$

However, there must exist $x_{1},x_{2}\in \mathbb {R}$ with $p(x_{1})\leq 0\leq p(x_{2})$. Further, for $k=0,\ldots n$, the polynomial $p$ is continuous as the composition of continuous functions $x\mapsto x^{k}$. Using Bolzano's root theorem, we can conclude the existence of some $x_{0}\in \mathbb {R}$ with $p(x_{0})=0$. Then the polynomial function as at least one root.

Range of Polynomials

Exercise (Range of Polynomials)

Let

$p:\mathbb {R} \to \mathbb {R} ,\ p(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\ldots +a_{1}x+a_{0}$

be a polynomial function with $a_{0},a_{1},\ldots ,a_{n-1},a_{n}\in \mathbb {R}$ and $a_{n}\neq 0$, and $n$ odd. Show $p(\mathbb {R} )=\mathbb {R}$.

Solution (Range of Polynomials)

Fall 1: $a_{n}>0$

For $x\neq 0$ it holds:

$\lim _{x\to \infty }p(x)=\lim _{x\to \infty }\underbrace {x^{n}} _{\to \infty }(\underbrace {a_{n}+a_{n-1}x^{-1}+\ldots +a_{1}x^{-n+1}+a_{0}x^{-n}} _{\to a_{n}>0})=\infty$

and

$\lim _{x\to -\infty }p(x)=\lim _{x\to -\infty }\underbrace {x^{n}} _{\to -\infty }(\underbrace {a_{n}+a_{n-1}x^{-1}+\ldots +a_{1}x^{-n+1}+a_{0}x^{-n}} _{\to a_{n}>0})=-\infty$

Since $p$ is continuous, it follows $p(\mathbb {R} )=\mathbb {R}$.

Fall 2: $a_{n}<0$

Similar to the first case.

Exercise: The existence of n-th order Roots

The intermediate value theorem also offers a possibility to check the existence of $n$-th order roots. In the chapter "Wurzel reeller Zahlen“ we have already proven this using the concept of nested intervals. Now we will see an alternative proof that uses the intermediate value theorem. Just a reminder: the $n$-th order root ${\sqrt[{n}]{\alpha }}$ for some positive number $\alpha$ is a real number $x$ with $x^{n}=\alpha$.

Exercise (Existence of n-th order Root)

Let $n\in \mathbb {N}$. Show that for all $\alpha >0$ there exists a positive number $x$ with $x^{n}=\alpha$.

Solution (Existence of n-th order Root)

It is the case that $x^{n}=\alpha$ if and only if $x^{n}-\alpha =0$. The desired number $x$ must therefore be e a positive zero of the polynomial $p:\mathbb {R} \to \mathbb {R}$ with $p(x)=x^{n}-\alpha$. First we observe that $p$, be a polynomial, is continuous on all. $\mathbb {R}$. Therefore $p$ is also continuous on the interval $[0,\alpha +1]$. Furthermore it holds

$p(0)=0^{n}-\alpha =-\alpha <0$

as well as

${\begin{aligned}p(1+\alpha )&=(1+\alpha )^{n}-\alpha \\[0.3em]&\quad {\color {Gray}\left\downarrow \ {\text{Bernoulli Inequality}}\right.}\\[0.3em]&\geq 1+n\alpha -\alpha \\&=1+\underbrace {(n-1)} _{\geq 0}\alpha \\&\geq 1\end{aligned}}$

Therefore $p(0)<0<p(1+\alpha )$. By the intermediate value theorem, $p$ has at least one zero in the interval $(0,1+\alpha )$. Because all numbers in $(0,1+\alpha )$ are positive, the zero must also be positive. Furthermore, the zero satisfies the equality $x^{n}=\alpha$.

Exercise: Solving an Equation

Both the intermediate value and root theorems can be used to justify the existence of a solution of a given equation. The equation is used to build a continuous function, on which we can apply either the root theorem or the intermediate value theorem.

Exercise (Solution to an Equation)

Show there exists only one $x\in \mathbb {R}$ with $\cos x=x$.

Solution (Solution to an Equation)

Functions $f(x)=\cos x$ and $g(x)=x$

Proof step: Show the equation has a solution

We will use the helping function

$h:\mathbb {R} \to \mathbb {R} ,\ h(x)=x-\cos x$

It is continuous since it is the difference of the two continuous functions $x\mapsto x$ and $x\mapsto \cos x$. Furthermore it holds

$h(0)=0-\cos 0=-1<0$

as well as

$h({\tfrac {\pi }{2}})={\tfrac {\pi }{2}}-\cos({\tfrac {\pi }{2}})={\tfrac {\pi }{2}}>0$

By the root theorem, there exists some ${\tilde {x}}\in \mathbb {R}$ with $h({\tilde {x}})={\tilde {x}}-\cos {\tilde {x}}=0$. For this ${\tilde {x}}$ it then holds $\cos {\tilde {x}}={\tilde {x}}$. This solves our equation.

Proof step: Show there exists exactly one solution

We will examine four cases:

Fall 1: $x\leq -{\tfrac {\pi }{2}}$

We have $x\leq -{\tfrac {\pi }{2}}<-1\leq \cos(x)$. Therefore there exists no $x$ with $\cos x=x$.

Fall 2: $-{\tfrac {\pi }{2}}<x\leq 0$

We have $x\leq 0<\cos x$. So there can be no $x$ satisfying $\cos x=x$.

Fall 3: $0<x<{\tfrac {\pi }{2}}$

We've already proven above that there exists some $x$ with $\cos x=x$. Since $x\mapsto x$ is strictly monotone increasing and $x\mapsto \cos x$ is strictly monotone falling on $0<x<{\tfrac {\pi }{2}}$, there can not exist some further $x$ such that $\cos x=x$.

Fall 4: ${\tfrac {\pi }{2}}\leq x$

We have $\cos x\leq 1<{\tfrac {\pi }{2}}\leq x$. So there can not exist any $x$ such that $\cos x=x$.

From both steps in the proof we can conclude that there exists exactly one $x\in \mathbb {R}$ with $\cos x=x$.